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A Concrete Approach to Abstract Algebra 

A Concrete Approach 
to Abstract 

W. W. Sawyer 

W. H. Freeman and Company 


The writing of this book, which was prepared while the 
author was teaching at the University of Illinois, as a 
member of the Academic Year Institute, 1957-1958, was 
supported in part by a grant from the National Science 

© Copyright 1959 by W. H. Freeman and Company, Inc. All 
rights to reproduce this book in whole or in part are reserved, 
with the exception of the right to use short quotations for review 
of the book. Printed in the United States of America. Library of 
Congress Catalogue Card Number: 59-10215 


Introduction 1 

1 The Viewpoint of Abstract Algebra 5 

2 Arithmetics and Polynomials 26 

3 Finite Arithmetics 71 

4 An Analogy Between Integers and Polynomials 83 

5 An Application of the Analogy 94 

6 Extending Fields 115 

7 Linear Dependence and Vector Spaces 131 

8 Algebraic Calculations with Vectors 157 

9 Vectors Over a Field 167 

10 Fields Regarded as Vector Spaces 185 

1 1 Trisection of an Angle 208 

Answers to Exercises 



The Aim of This Book 
and How to Read It 

At the present time there is a widespread desire, 
particularly among high school teachers and engineers, 
to know more about “modern mathematics.” Institutes 
are provided to meet this desire, and this book was 
originally written for, and used by, such an institute. 
The chapters of this book were handed out as mimeo- 
graphed notes to the students. There were no “lectures”; 
I did not in the classroom try to expound the same mate- 
rial again. These chapters were the “lectures.” In the 
classroom we simply argued about this material. Ques- 
tions were asked, obscure points were clarified. 

In planning such a course, a professor must make a 
choice. His aim may be to produce a perfect mathemat- 
ical work of art, having every axiom stated, every con- 
clusion drawn with flawless logic, the whole syllabus 
covered. This sounds excellent, but in practice the result 
is often that the class does not have the faintest idea of 
what is going on. Certain axioms are stated. How are 
these axioms chosen? Why do we consider these axioms 
rather than others? What is the subject about? What is 
its purpose? If these questions are left unanswered, stu- 
dents feel frustrated. Even though they follow every 


A Concrete Approach to Abstract Algebra 

individual deduction, they cannot think effectively about 
the subject. The framework is lacking; students do not 
know where the subject fits in, and this has a paralyzing 
effect on the mind. 

On the other hand, the professor may choose familiar 
topics as a starting point. The students collect material, 
work problems, observe regularities, frame hypotheses, 
discover and prove theorems for themselves. The work 
may not proceed so quickly ; all topics may not be 
covered; the final outline may be jagged. But the student 
knows what he is doing and where he is going; he is 
secure in his mastery of the subject, strengthened in con- 
fidence of himself. He has had the experience of discover- 
ing mathematics. He no longer thinks of mathematics 
as static dogma learned by rote. He sees mathematics 
as something growing and developing, mathematical 
concepts as something continually revised and enriched 
in the light of new knowledge. The course may have cov- 
ered a very limited region, but it should leave the student 
ready to explore further on his own. 

This second approach, proceeding from the familiar 
to the unfamiliar, is the method used in this book. 
Wherever possible, I have tried to show how modern 
higher algebra grows out of traditional elementary alge- 
bra. Even so, you may for a time experience some feeling 
of strangeness. This sense of strangeness will pass; there 
is nothing you can do about it; we all experience such 
feelings whenever we begin a new branch of mathemat- 
ics. Nor is it surprising that such strangeness should be 
felt. The traditional high school syllabus — algebra, ge- 
ometry, trigonometry — contains little or nothing dis- 
covered since the year 1650 a.d. Even if we bring in 
calculus and differential equations, the date 1750 a.d. 
covers most of that. Modern higher algebra was de- 
veloped round about the years 1900 to 1930 a.d. Anyone 



who tries to learn modern algebra on the basis of tradi- 
tional algebra faces some of the difficulties that Rip Van 
Winkle would have experienced, had his awakening 
been delayed until the twentieth century. Rip would 
only overcome that sense of strangeness by riding around 
in airplanes until he was quite blase about the whole 

.. Some comments on the plan of the book may be 
helpful. Chapter 1 is introductory and will not, I hope, 
prove difficult reading. Chapter 2 is rather a long one. 
In a book for professional mathematicians, the whole 
content of this chapter would fill only a few lines. I 
tried to spell out in detail just what those few lines would 
convey to a mathematician. Chapter 2 was the result. 
The chapter contains a solid block of rather formal 
calculations (pages 50-56). Psychologically, it seemed a 
pity to have such a block early in the book, but logically 
I did not see where else I could put it. I would advise 
you not to take these calculations too seriously at a first 
reading. The ideas are explained before the calculations 
begin. The calculations are there simply to show that 
the program can be carried through. At a first reading, 
you may like to take my word for this and skip pages 
50-56. Later, when you have seen the trend of the whole 
book, you may return to these formal proofs. I would 
particularly emphasize that the later chapters do not in 
any way depend on the details of these calculations — 
only on the results. 

The middle of the book is fairly plain sailing. You 
should be able to read these chapters fairly easily. 

I am indebted to Professor Joseph Landin of the 
University of Illinois for the suggestion that the book 
should culminate with the proof that angles cannot be 
trisected by Euclidean means. This proof, in chapter 11, 
shows how modern algebraic concepts can be used to 


A Concrete Approach to Abstract Algebra 

solve an ancient problem. This proof is a goal toward 
which the earlier chapters work. 

I assume, if you are a reader of this book, that you 
are reasonably familiar with elementary algebra. One 
important result of elementary algebra seems not to be 
widely known. This is the remainder theorem. It states 
that when a polynomial f(x) is divided by x — a, the 
remainder is f(a). If you are not familiar with this 
theorem and its simple proof, it would be wise to review 
these, with the help of a text in traditional algebra. 

Chapter i 

The Viewpoint of Abstract 

There are two ways in which children do arithmetic 
— by understanding and by rote. A good teacher, cer- 
tainly in the earlier stages, aims at getting children to 
understand what 5—2 and 6X8 mean. Later, he 
may drill them so that they will answer “48” to the 
question “Eight sixes?” without having to draw eight 
sets of six dots and count them. 

Suppose a foreign child enters the class. This child 
knows no arithmetic, and no English, but has a most 
retentive memory. He listens to what goes on. He notices 
that some questions are different from others. For in- 
stance, when the teacher makes the noise “What day is 
it today?” the children may make the noise “Monday” 
or “Tuesday” or “Wednesday” or “Thursday” or “Fri- 
day.” This question, he notices, has five different an- 
swers. There are also questions with two possible answers, 
“Yes” and “No.” For example, to the question “Have 
you finished this sum?” sometimes one, sometimes the 
other answer is given. 

However, there are questions that always receive the 
same answer. “Hi” receives the answer “Hi.” “Twelve 


A Concrete Approa ch to Abstract Algebra 

twelves?” receives the answer “A hundred and forty- 
four” — or, at least, the teacher seems more satisfied 
when this response is given. Soon the foreign child might 
learn to make these responses, without realizing that 
“Hi” and “144” are in rather different categories. 

Suppose that the foreign child comes to school after 
the children in his class have finished working with blocks 
and beads. He sees 12 X 12 = 144 written and hears it 
spoken, but is never present when 12 is related to the 
counting of twelve objects. 

One cannot say that he understands arithmetic, but 
he may be top of the class when it comes to reciting the 
multiplication table. With an excellent memory, he may 
have complete mastery of formal, mechanical arithmetic. 

We may thus separate two elements in arithmetic, 
(i) The formal element — this covers everything the for- 
eign child can observe and learn. Formal arithmetic is 
arithmetic seen from the outside, (ii) The intuitive ele- 
ment — the understanding of arithmetic, its meaning, 
its connection with the actual world. This understand- 
ing we derive by being part of the actual universe, by 
experiencing life and seeing it from the inside. 

For teaching, both elements of arithmetic are neces- 
sary. But there are certain activities for which the formal 
approach is helpful. In the formalist philosophy of math- 
ematics, a kind of behaviorist view is taken. Instead of 
asking “How do mathematicians think?” the formalist 
philosophers ask “What do mathematicians do?” They 
look at mathematics from the outside: they see mathe- 
maticians writing on paper, and they seek rules or laws 
to describe how the mathematicians behave. 

Formalist philosophy is hardly likely to provide a full 
picture of mathematics, but it does illuminate certain 
aspects of mathematics. 

A practical application of formalism is the design of 
all kinds of calculating machines and automatic appli- 

T he Viewpoint of Abstract Algebra 


ances. A calculating machine is not expected to under- 
stand what 71 X 493 means, but it is expected to give 
the right answer. A fire alarm is not expected to under- 
stand the danger to life and the damage to property 
involved in a fire. It is expected to ring bells, to turn on 
sprinklers, and so forth. There may even be some con- 
nection between the way these mechanisms operate and 
the behavior of certain parts of the brain. 

One might say that the abstract approach studies 
what a machine is, without bothering about what it 
is for. 

Naturally, you may feel it is a waste of time to study 
a mechanism that has no purpose. But the abstract ap- 
proach does not imply that a system has no meaning 
and no use; it merely implies that, for the moment, we 
are studying the structure of the system, rather than its 

Structure and purpose are in fact two ways of classify- 
ing things. In comparing a car and an airplane, you 
would say that the propeller of an airplane corresponds 
to the driving wheels of a car if you are thinking in terms 
of purpose; you would however say that the propeller 
corresponds to the cooling fan if you are thinking in 
terms of structure. 

Needless to say, a person familiar with all kinds of 
mechanical structures — wheels, levers, pulleys, and so 
on — can make use of that knowledge in inventing a 
mechanism. In a really original invention, a structure 
might be put to a purpose it had never served before. 

Arithmetic Regarded as a Structure 

Accordingly, we are going to look at arithmetic from 
the viewpoint of the foreign student. We shall forget 
that 12 is a number used for counting, and that + and 
X have definite meanings. We shall see these things 


A Concrete Approach to Abstract Algebra 

purely as signs written on the keys of a machine. 

Stimulus: 12 X 12. 

Response: 144. 

Our calculating machine would have the following 
visible parts: 

(i) A space where the first number is recorded. 

(ii) A space for the operations +, X, — , 

(iii) A space for the second number. 

These constitute the input. 

The output is the answer, a single number. 

Playing around with our machine, we would soon 
observe certain things. Order is important with -r- and 
— . Thus 6 -T- 2 gives the answer 3, while 2 -r- 6 gives 
the answer 1/3. But order is not important with + and 
X • Thus 3 + 4 and 4 + 3 both give 7 ; 3 X 4 and 4X3 
both give 12. 

We have the commutative laws: a b = £ + a, a X. b = 
b X a. (Or ab = ba, with the usual convention of leav- 
ing out the multiplication sign.) 

Commutativity is not something that could have been 
predicted in advance. Since 6 -S- 2 is not the same as 
2 "7" 6, we could not say, for any sign S, that 

a S b — b S a. 

Some comment may be made here on the symbol S. 
In school algebra, letters usually stand for numbers. In 
what we are doing, letters stand for things written on the 
keys of machines. The form a S b covers, for example, 

a “times” b, 
a “plus” b, 
a “minus” b, 
a “over” b, 
a “to the power” b, 
a “-th root of” b, 
a “-’s log to base” b , 

The Viewpoint of Abstract Algebra 9 

as well as many more complicated ways of combining 
a and b that one could devise. 

Commutativity, then, is something we may notice 
about a machine. It is one example of the kind of 
remark that can be made about a machine. 

Ordinary arithmetic has one property that is incon- 
venient for machine purposes: it is infinite. If we make 
a calculating machine that goes up to 999,999 we 
are unable to work out, say, 999,999 + 999,999 or 
999,999 X 999,999 by following the ordinary rules for 
operating the machine. 

We can consider a particular calculating machine that 
is very much simpler, and that avoids the trouble of 
infinity. This machine will answer any question appropri- 
ate to its system. It deals with a particular part or aspect 
of arithmetic. 

If two even numbers are added together, the result is 
an even number. If an even number is added to an odd 
number, the result is odd. We may, in fact, write 

Even -f- Even = Even 
Even + Odd = Odd 
Odd + Odd = Even. 

Similarly, there are multiplication facts, 

Even X Even = Even 
Even X Odd = Even 
Odd X Odd = Odd. 

Here we have a miniature arithmetic. There are only 
two elements in it, Even and Odd. Let us abbreviate, 
writing A for Even, B for Odd. Then 

A A = A A XA = A 

A + B = B AX B = A 

B + A = B B X A = A 

B + B = A B X B = B 


A Concrete Approach to Abstract Algebra 

which may be written more compactly as 

A B 

, a Fa b 

+ B B A 







Our foreign student would have no reason for regard- 
ing A and B in the tables above as any different from 
0, 1, 2, 3, • • • , in the ordinary addition and multiplica- 
tion tables. He might think of it as “another arithmetic.” 
He does not know anything about its meaning. What 
can he observe about its structure? Does it behave at all 
like ordinary arithmetic? In actual fact, the similarities 
are very great. I shall only mention a few of them at 
this stage. 

Both addition and multiplication are commutative in 
the A, B arithmetic. For instance, A + B = B + A and 
A X B = B X A. 

In ordinary arithmetic the number zero occurs. We 
know the meaning of zero. But how could zero be iden- 
tified by someone who only saw the structure of arith- 
metic? Quite easily, for there are two properties of zero 
that single it out. First, when zero is added to a number, 
it makes no difference. Second, whatever number zero 
is multiplied by, the result is always zero. 


x + 0 = x, 
x • 0 = 0. 

Is there a symbol in the A, B arithmetic that plays 
the role of zero? It makes a difference when B is added : 
A + B is not A, nor is B + B the same as B. A is the 
only possible candidate, and in fact A passes all the tests. 
When you add A, it makes no difference ; when anything 
is multiplied by A, you get A. 

Is there anything that corresponds to 1? The only dis- 

The Viewpoint of Abstract Algebra 


tinguishing property I can think of for 1 is that mul- 
tiplication by 1 has no visible effect: 

*•1 = x. 

In the A, B arithmetic, multiplication by B leaves 
any symbol unchanged. So B plays the part of 1. 

This suggests that we might have done better to 
choose O (capital o) as a symbol instead of A and I as 
a symbol instead of B, because O looks like zero, and I 
looks rather like 1 . 

Our tables would then read 

0 + 0 = 0 
0 + 1=1 
1+0 = 1 
1 + 1=0 

0X0 = 0 
1X0 = 0 
I X I = I 

Now this looks very much like ordinary arithmetic. 
In fact, the only question that would be raised by some- 
body who thought I stood for 1 and O for zero would be, 
“Haven’t you made a mistake in writing I + I = O?” 
All the other statements are exactly what you would 
expect from ordinary arithmetic. 

The tables of this “arithmetic” are 




O I 

0 I 

1 o 

O I 

o o 

O I 

We arrived at the tables above by considering even 
and odd numbers. But we could arrive at the same pat- 
tern without any mention of numbers. 

Imagine the following situation. There is a narrow 
bridge with automatic signals. If a car approaches from 
either end, a signal “All clear — Proceed” is flashed on. 
But if cars approach from both ends, a war nin g signal 

12 A Concrete Approach to Abstract Algebra 

is flashed, and the car at, say, the north end is instructed 
to withdraw. 

In effect, the mechanism asks two questions: “Is a car 
approaching from the south? Is one approaching from 
the north?” The answers to these questions are the input, 
the stimulus. The output, the response of the mechanism, 
is to switch on an appropriate signal. 

For the all-clear signal the scheme is as follows. 

Should all-clear signal be flashed ? 

Car from north? 

No Yes 

Car No 


south? Yes 

For the warning signal the scheme is as follows. 

Should warning signal be flashed? 

Car from north? 

No Yes 

Car No 


south? Yes 

If you compare these tables with the earlier ones, you 
will see that they are exactly the same in structure. 
“No” replaces “O,” “Yes” replaces “I”; “all clear” is 
related to +, “warning signal” to X. 

One could also realize this pattern by simple electrical 

O O 0 0 



The Viewpoint of Abstract Algebra 


If you had this machine in front of you, you would not 
know whether it was intended for calculations with even 
and odd numbers, or for traffic control, or for some other 

When the same pattern is embodied in two different 
systems, the systems are called isomorphic. In our exam- 
ple above, the traffic control system is isomorphic with 
the arithmetic of Even and Odd. The same machine 
does for both. 

Isomorphism does not simply mean that there is some 
general resemblance between the two systems. It means 
that they have exactly the same pattern. Our example 
above shows this exact correspondence. Wherever “O” 
occurs in one system, “No” occurs in the other; wher- 
ever “I” occurs in one system, “Yes” occurs in the other. 

The statements, “these two systems are isomorphic” 
and “there is an isomorphism between them,” are two 
different ways of saying the same thing. To prove two 
systems isomorphic, you must demonstrate a correspond- 
ence between them, like the one in our example. 

The study of structures has two things to offer us. 
First, the same structure may have many different re- 
alizations. By studying the single structure, we are simul- 
taneously learning several different subjects. 

Second, even though we have only one realization of 
our structure in mind, we may be able to simplify our 
proofs and clarify our understanding of the subject by 
treating it abstractly — that is to say, by leaving out 
details that merely complicate the picture and are not 
relevant to our purpose. 

Our Results Considered Abstractly 

So far we have been concrete in our approach. That is, 
we have been talking of things whose meaning we under- 


A Concrete Approach to Abstract Algebra 

stand — numbers, Even and Odd, Yes and No. This is, 
of course, desirable from a teaching point of view, to 
avoid an unbearable sense of strangeness. 

Now let us look at what we have found purely in terms 
of pattern, of structure, and without reference to any 
particular interpretation or application it may have. 
That is, we return strictly to the point of view of the 
foreign student. What can we say? 

Well, first of all, we can recognize what belongs to 
the subject. Arithmetic deals with 0, 1,2, 3, 4, 5, 6, 7, 
8, 9. 7 is an element of arithmetic; “Hi” is not. “O” 
and “I” were elements used in our miniature arithmetic. 
“Yes” and “No” were elements in the traffic problem. 
The various positions of the switches were elements in 
the electrical mechanism. 

So, first of all, our subject deals with a certain set of 
recognizable objects. Then we have a certain procedure 
with these objects. If we take the electrical machine 
marked +, and set the first switch to I, the second to O, 
the machine gives us the response I. We say I + O is I. 
In the same way, if the teacher asks “3 + 4?” the chil- 
dren respond “7.” 

We may call adding and multiplying operations. A 
machine might be devised to do many other operations 

Thus in arithmetic we specify the objects 3, 4 and 
the operation “add” (+). The machine or the class 
gives us another object, 7, as a response. 

We can list the things we noted earlier about arith- 

(1) Arithmetic deals with a certain set of objects. 

(2) Given any two of these objects a, b, another object 
called their sum is uniquely defined. If c is the sum 
of a and b, we write c = a + b. 

The Viewpoint of Abstract Algebra 


(3) In the same way a product k is defined. We write 
k = a X b or k = a- b. 

(4) a + b and b + a are the same object. 

(5) a-b and b-a are the same object. 

(6) There is an object 0 such that a + 0 = a and 
<3-0 = 0 for every a. 

(7) There is an object 1 such that a • 1 = a for every a. 
These are not all the things that could be said about 

arithmetic. We have not mentioned the associative laws, 
{a + b) + c = a + {b + c), a-(b-c) = ( a-b)-c; the dis- 
tributive law, a(x +jy) = ax -f- ay; nor anything about 
subtraction and division. 

However, suppose we agree that statements (1) 
through (7) are enough to think about for the moment. 
We might ask, “Is ordinary arithmetic the only struc- 
ture with these properties? If not, what is the smallest 
number of objects with which this structure can be 

We already have the answer to both questions. Arith- 
metic is not the only structure satisfying statements (1) 
through (7). The smallest structure consists of the ob- 
jects O, I with the tables for + and X given earlier. 
(We are assuming that 0 and 1 are distinct objects.) 


1. Let O stand for “any number divisible by 3,” I for “any 
number of the form 3 n + 1,” and II for “any number of the 
form 3 n + 2.” Can one say to what class a + b will belong 
if one knows to what classes a and b belong? And the product ab? 
If so, form tables of addition and multiplication, as we did 
with the tables for Even and Odd. Do statements (1) through 
(7) apply to this topic? 

16, A Concrete Approach to Abstract Algebra 

2. The same as question 1, but with the classes O, I, II, III, 
IV for numbers of the forms 5 n, 5« + 1, 5n + 2, 5n + 3, 
5n + 4. 

3. Continue the inquiry for other numbers, 4, 6, 7, • • • , 
replacing 3 and 5 of questions 1 and 2. Do you notice any 
differences between the results for different numbers? 

4. An arithmetic is formed as follows. The only permitted 
objects are 0, 2, 4, 6, 8. When two numbers are added or 
multiplied, only the last digit is recorded. For example, in 
ordinary arithmetic 6 + 8 = 14 with last digit 4. In this arith- 
metic 6 + 8 = 4. Normally 4 X 8 = 32 with last digit 2. So 
here 4X8 = 2. Write out the addition and multiplication 
tables. Do statements (1) through (7) apply here? This arith- 
metic contains five objects, as did the arithmetic of question 2. 
Are the arithmetics isomorphic? 

5. Calculate the powers of II, of III, and of IV in the 
arithmetic of question 2. 

6. Are subtraction and division possible in the arithmetic of 
question 2? Do they have unique answers? What about the 
arithmetics you studied under question 3? 

7. In the arithmetic of question 2, which numbers are per- 
fect squares? Which numbers are prime? Does this arithmetic 
have any need of (i) negative numbers, (ii) fractions? 

Two Arithmetics Compared 

There is a certain stage in the learning of arithmetic 
at which the only operations known are addition, sub- 
traction, multiplication, and division. The child has not 
yet met V2, but is familiar with whole numbers and 
fractions. I am not sure whether it would be so in cur- 
rent educational practice, but we shall suppose the child 
knows about negative numbers. 

The charm of this stage of knowledge is that every 
question has an answer. You must not, of course, ask for 
division by zero, but, apart from this reasonable restric- 

The Viewpoint of Abstract Algebra 


tion, if you are given any two numbers you can add, 
subtract, multiply, or divide and reach a definite answer. 

The body of numbers known to a child at this stage 
are referred to as the rational numbers. The rational 
numbers comprise all numbers of the form p/q, where 
p and q are whole numbers (positive or negative) ; p can 
also be zero but q must not. Since q can be 1, we have 
not excluded the whole numbers themselves. 

The operations the child knows at this stage we may 
call the rational operations. A rational operation is any- 
thing that can be done by means of addition, subtrac- 
tion, multiplication, and division, each used as often as 
you like. For instance, 

(* + D(y - h) 


is the result of a rational operation on x and y. Note, 
however, that the process must finish. A child in grade 
school is not expected to cope with an expression like 

1 + 

2 + 

2 + 

2 + 

2 + 

and so on forever. This expression, in a certain sense, 
represents V2. The study of unending processes belongs to 
analysis: we exclude any such idea from algebra. 

To sum up: There is a stage when a child sees arith- 
metic as consisting of rational operations on rational 
numbers. At this stage, every question has an answer, 
every calculation can be carried out. 

Now we consider another arithmetic. On an island in 


A Concrete Approach to Abstract Algebra 

the Pacific, a sociological experiment is being performed. 
A child goes to school and learns the addition and multi- 
plication tables. This sounds quite normal. But the tables 
he learns are the following. 



+ 2 



0 12 3 4 

0 12 3 4 

1 2 3 4 0 

2 3 4 0 1 

3 4 0 1 2 

4 0 12 3 



X 2 



0 12 3 4 
0 0 0 0 0 
0 12 3 4 
0 2 4 1 3 
0 3 14 2 
0 4 3 2 1 

You will recognize this arithmetic from a question in 
the preceding section. 0, 1, 2, 3, 4 are the possible re- 
mainders on division by 5. 

But the child has no such background. He is simply 
taught the tables above. If he says 4 X 2 = 3, he is 
rewarded. If he says 4X2 equals anything else, he is 
punished. Now we compare arithmetic as experienced by 
this child with ordinary arithmetic as learned by an 
ordinary child. 

To begin with, both children would accept the fol- 
lowing statements. 

(/) You can add any two numbers a and b, and there is 
only one correct answer. 

(2) You can multiply any two numbers a and b and there 
is only one correct answer. 

(3) a + b — b + a, for all numbers a , b. 

( 4 ) ab = ba , for all numbers a, b. 

(5) a + (b + c) = (a + b) + c, for all numbers a , b, c. 

( 6 ) a(bc) = ( ab)c , for all numbers a, b, c. 

(7) a(b + c) = ab + ac, for all numbers a, b, c. 

I do not know if the children would be able to prove 
that all these were so, but at least they would be able 
to take various particular cases, and admit they could 
not find any instance in which any of these statements 
was false. 

The Viewpoint of Abstract Algebra 


Statement (5), for instance, in ordinary arithmetic, 
expresses the fact that when you are adding, say, the 
numbers 7, 11, and 13 it does not matter whether you 

7 + 11 = 18, and 18 + 13 = 31 

11 + 13 = 24, and 7 + 24 = 31. 

The intermediate steps look quite different, but they 
lead to the same final result. 

In the miniature arithmetic, an example would be 
adding 2, 3, and 4. You could either say 

2 + 3 = 0, and 0 + 4 = 4 

3 + 4 = 2, and 2 + 2=4, 

the final answer being 4 either way. 

Statement ( 6 ) expresses the fact that you can work 
out 7 X 11 X 13, by writing 

7X11= 77, and 77 X 13 = 1,001 
or equally well by writing 

11 X 13 = 143, and 7 X 143 = 1,001. 

Statement (7) expresses our experience that we can 
work out 4 X (2 + 5) equally well as 

4 X 7 = 28 

or as 

4X2 + 4X5 = 8 + 20 = 28. 

Corresponding procedures apply in the miniature 
arithmetic, though the results look strange to us. 

If we had to work sums in the miniature arithmetic, 
there would be many of our habits that we could carry 
over and use to obtain correct results. In fact, statements 
(3) to (7) embody a very large part of the rules that we 
follow, consciously or unconsciously, in doing arithmetic 
or algebra. 


A Concrete Approach to Abstract Algebra 

Subtraction and Division 

Our seven statements above make no mention of sub- 
traction or division. When we learn arithmetic, 7 — 4 
is probably explained as “4 and what are 7?” This is, 
in everything except language, an invitation to solve the 
equation 4 + * = 7. Further, grade school teachers have 
a strong prejudice to the effect that this equation has 
only one solution, x — 3. 

The formal statement ( 8 ) below therefore contains 
nothing more than our own infant experiences. 

( 8 ) (i) For all a and b, the equation a + x = b has a 
solution, (ii) The equation has only one solution, 
(iii) This solution is called b — a. 

Here (i) and (ii) make statements that can be tested by 
taking particular numbers for a and b. Statement (iii) 
merely explains what we understand by the new sym- 
bol, — , that we have just brought in. It does not require 
testing or proof. It shows us, however, how to find, say, 
2 — 3 in the miniature arithmetic. 2 — 3 is the number 
that satisfies 3 + x — 2. In the addition table, we must 
look along the row opposite 3, until we find the num- 
ber 2. We find it under 4, and only under 4. 3 -j- 4 = 2, 
and no other number will do in place of 4. So 2 — 3 = 4 
is correct, and is the only correct answer. 

Question: What does the requirement, that a + x = b 
shall have one and only one solution for all a and b, tell us 
about the rows of the addition table? 

When you subtract a number from itself, the answer 
is zero. We might express this in the statement: a — a 
has a fixed value, independent of what a is; this value 
is called 0. 

As a — a = 0 means the same thing as a = a + 0, we 
can equally well put this statement in the following 
form (that we have already met). 

The Viewpoint of Abstract Algebra 


(9) There is a number 0 such that, for every a, a + 0 = a. 
There can of course only be one such number; other- 
wise a x = a would have more than one solution, 
which would contradict part (ii) of statement ( 8 ). 

Now we come to division. As children we meet divi- 
sion in much the same way as subtraction. “4 and what 
is 12?” is replaced by “4 times what is 12?” We might 
begin to write a statement, on the lines of (8), that 
ax — b has a solution, and only one solution, whatever 
a and b. But this would overlook the fact that 

0 -* = 0 

is satisfied by every number x, so that this equation has 
more than one solution, while 

0-x = 1 

is not satisfied by any number x. 

Apart from this point, there is no difficulty in giving 
a formal statement of our experiences with division. 

(10) For all a and b, provided however that a is not 0, 
(i) ax = b has a solution, (ii) This equation has 
only one solution, (iii) The solution is called 
b -§- a or b/a. 

Just as we can find a — a to be 0 without knowing 
what number a is, we know in ordinary arithmetic that 
a -f a or a/a is 1, without needing to know what a is, 
except of course that a is not 0. 

So, as statement (8) was followed by (9), statement 

(10) is followed by (11). 

(11) There is a number 1 such that, for every a, a • 1 = a. 
If you will now test statements ( 8 ), (9), (10), (11) for 

the miniature arithmetic, you will find that all of them 
work for it too. 

This is quite remarkable. Within the set 0, 1, 2, 3, 4, 
without having to introduce any fresh numbers (like 
negative numbers or fractions in ordinary arithmetic), 


A Concrete Approach to Abstract Algebra 

we can add, subtract, multiply, and divide to our heart’s 

For instance, in the miniature arithmetic, simplify 

(1 + fHt-l) 

3 3. 

UT 4 

We can get rid of the fractions in the numerator and 
denominator immediately: 

| = 1 V 2 = 3, 
f = 2 -h 3 = 4, 
f = 3 -f- 4 = 2, 
f = 3 -5- 2 = 4. 

The fraction is simplified to 

(3 + 4). (4 -2) _ 2-2 _ 

4-2 2 

There are many different ways in which this expres- 
sion could be simplified. In the denominator, for in- 
stance, we could say 3/2 — 3/4 = 3/4; and then divi- 
sion by 3/4 is the same as multiplication by 4/3. This 
would give 

i • (i + 1) • (t - !), 

which can still be simplified in several ways. But how- 
ever we proceed, we shall always arrive at the answer 2. 

You may have noticed that 2 — 3 = 4, 3-1-2 = 4. 
This shows that, for x = 2, x — 3 = 3/x. Are there any 
other solutions of this equation? We have 


Multiply by x: 

x 2 — 3x = 3. 

Subtract 3. Since 0 — 3 = 2, this gives 

x 2 — 3x + 2 = 0. 

(* — l)(x — 2) = 0. 

The Viewpoint of Abstract Algebra 


So x — 1 and * = 2 are solutions. Could there be any 
more solutions? To show there are not we need to ob- 
serve (72). 

(72) ab = 0 only if a = 0 or b = 0. In words, a product 
is zero only if a factor is zero. 

Above we had (x — 1)(* — 2) = 0. Either 

* — 1= 0 or * — 2 = 0; 

* = 1 or * = 2. 

Thus quadratics can be solved by factoring exactly as 
in ordinary arithmetic. They can also be solved by 
completing the square. For example, consider 

* 2 + x = 2. 

To complete the square for * 2 + ax, we add {a/2) 2 to 
each side. In our equation a = 1, so a/2 = 3, since 
2X3 = 1. We must add to each side 3 2 , that is, 4. Thus 

* 2 + * + 3 2 = 2 + 4 = 1, 

(* + 3) 2 = 1. 

Next we have to take the square root. I 2 = 1 and also 
4 2 = 1. (Note that 4 = 0 — 1, so that ±1 is the same 
as 1 or 4.) Thus 

* + 3 = 1 or * + 3 = 4; 

* = 3 or * = 1. 

One can test by substituting these values in the orig- 
inal equation that they actually are roots. 

This arithmetic will be referred to as “the arithmetic 
modulo 5.” The number 5 in this title indicates that we 
are dealing with 0, 1, 2, 3, 4, the possible remainders 
on division by 5. 

In the same way, the arithmetic of Even and Odd, 
with elements 0, 1 is called “arithmetic modulo 2.” 
(On division by 2, an even number leaves remainder 0, 
an odd number 1.) 


A Concrete Approach to Abstract Algebra 

In earlier exercises, you were invited to study the 
arithmetics modulo 3, modulo 4, modulo 6, modulo 7, 
and so on. 


1. Make a table of squares, cubes, and fourth powers 
modulo 5. Solve the equations x 2 = 1, x 3 = 1, x* = 1 in this 

2. Find (x + y) 5 modulo 5. 

3. Divide x 2 + 1 by x -f- 2, modulo 5. Has x 2 + 1 =0 
any solutions in this arithmetic? What are they? 

4. In the text we solved x 2 + x = 2, modulo 5. This equa- 
tion may be written x 2 + x + 3 = 0. What are the factors of 
x 2 + x + 3? 

5. Find by trial, by completing the square, or by any 
other method, the solutions of the following equations in the 
arithmetic modulo 5 : (i) x 2 + 2x + 2 = 0, (ii) x 2 + 3x + 1 =0, 
(iii) x 2 + x -j- 4 = 0, (iv) x 2 + 4 = 0. What are the factors of 
the quadratic expressions that occur in the equations above? 

6. Divide X s + 2x 2 + 3x + 4 by x — 2 in the arithmetic 
modulo 5. 

7. Does the remainder theorem hold in the arithmetic 
modulo 5? 

8. Does the equation (x — 3) 2 = 0 have any solution other 
than x — 3 in the arithmetic modulo 4? 

9. In the arithmetic modulo 6 calculate the values of 
x 2 + 3x + 2 for x = 0, 1, 2, 3, 4, 5. How many roots does the 
quadratic equation x 2 + 3x + 2 = 0 have in this arithmetic? 

10. All the statements (7) through (72) are true in the 
arithmetic modulo 5. Which of them hold in (i) the arithmetic 
modulo 4, (ii) the arithmetic modulo 6? 

11. Can it be proved that a quadratic equation has at most 
two roots (i) in an arithmetic where statements (7) through (9) 

The Viewpoint of Abstract Algebra 


and (77) only are known to hold? (ii) in an arithmetic where 
statements (7) through (72) are known to hold? 

12. In the arithmetic modulo 5 are there any quadratics 
* 2 + px + q (i) that have no solutions when equated to zero? 
(ii) that cannot be split into factors of the form (x + «)(* + b)? 

13. In the arithmetic modulo 5 the equation x 3 = 2 has the 
solution 3. Has it any other solutions? Divide x 3 — 2 by x — 3. 
Has the resulting quadratic any factors? 

Chapter 2 

Arithmetics and Polynomials 

We have now met three kinds of arithmetic. Our or- 
dinary arithmetic is the first kind. It deals with numbers 
0, 1, 2, • • • , that go on forever. 

The arithmetic modulo 5 is the second kind. It con- 
tains only 0, 1, 2, 3, 4, but in spite of this it is remarkably 
like ordinary arithmetic. I can still ask you quite con- 
ventional questions in algebra — to multiply expressions, 
to do long division, to solve a quadratic, to factor a 
polynomial, to prove the remainder theorem. 

The third type is shown by arithmetic modulo 4 or 
modulo 6. It diverges still further from ordinary arith- 
metic. A quadratic may have more than two roots; 
still more striking, division ceases to be possible. In 
modulo 6 arithmetic, 3-r 2 has no answer, while 4 + 2 
has the answers 2 and 5. However, some similarities to 
ordinary arithmetic remain. We can still multiply with- 
out restriction. We can divide by 1 and 5, and this means 
that we can divide by x — a or 5x — a. The remainder 
theorem, that a polynomial f(x) on division by x — a 
leaves the remainder /(a), still makes sense and is true. 

It will be helpful in considering these arithmetics, and 
other structures that we shall meet, to tabulate their 
properties. On the left side of our table, we write our 

Rational Natural 

Statements numbers numbers Integers Modulo 5 Modulo 6 


+ + + + + + + C 1 + o + + 






^ S3 

V3 -S3 


S3 g 


-S3 ^3 




<43 XS 
<U <u 
X5 fl 

** *S M + 53 






~S X 

S3 44 c/3 





-S3 -S3 

x to 

<u .53 
X „ X 

flj <3 CJ 

S3 S3 S3 S3 S3 S3 S3 -S3 O -S3 


A Concrete Approach to Abstract Algebra 

statements (7) through (72). Across the top of the table, 
we write the names of the structures we plan to “test.” 
If a structure satisfies the tests, or statements, we enter 
a plus sign in the proper column. If a structure fails to 
satisfy a test, or statement, we enter a zero. It is often 
the property that is lacking that gives a peculiar flavor. 

Across the top of our table we have the following struc- 
tures listed: (i) The rational numbers; (ii) The natural 
numbers 0, 1, 2, • • • ; (iii) The integers 0, ±1, ±2, • ; 

(iv) The arithmetic modulo 5 ; (v) The arithmetic mod- 
ulo 6. 

In future, we shall define various types of structures by 
saying which tests are passed. A table of this kind gives 
a convenient way of recording definitions and of classify- 
ing any particular structure. 

We shall give one such definition straight away. Our 
table shows two structures that make exactly the same 
score — the rational numbers and the arithmetic mod- 
ulo 5. Now we have had several examples to show that 
you can work modulo 5 very much as you do in ordinary 
arithmetic. We therefore introduce a name to express 
this kind of similarity. 

definition. Any structure that passes all the tests (7) 
through (12) is called a field. 

It is hard to hold all the twelve tests in mind at once, 
and a rather looser explanation may be easier to remem- 
ber. A field is a structure in which you can add, subtract, 
multiply, and divide, and these operations behave very 
much as they do in elementary arithmetic. Tests (7) 
through (72) make precise what I mean by “behave 
very much alike.” 

It may be well to collect together the twelve tests, 
and to state them in a way that we can use generally. 
Several of them were stated above in terms of the child 
in the Pacific Island. 

Arithmetics and Polynomials 


Every structure we consider contains elements. We are 
not concerned with what these elements are; they may 
be marks on paper, sounds of words, physical objects, 
parts of a calculating machine, thoughts in the mind. 
We also have operations +, X or +, ♦. These opera- 
tions need not have any connection with addition and 
multiplication in arithmetic, other than the purely for- 
mal resemblance required by the tests below. We think 
again of our calculating machine, with two spaces for 
elements a, b ; one space for a sign + or • ; and a space 
for the answer. We understand by a + b or a-b what 
appears in the answer space — regardless of the internal 
mechanism of the calculating machine. 

The following twelve statements will henceforth be 
referred to as the axioms for a field. 

(7) To any two elements a, b and the operation +, there 
corresponds a uniquely defined element c. We write 
c = a + b. 

(2) To any two elements a, b and the operation there 
corresponds a uniquely defined element d. We write 
d = a-b. 

(3) a + b = b + a, for all elements a , b. 

(4) a-b = b-a,for all elements a, b. 

(5) a + (b + c) = (a + b) + c, for all elements a, b, c. 

(6) a-(b-c) = ( a-b)-c , for all elements a, b, c. 

(7) a-(b + c) = (a-b) -j- ( a-c ), for all elements a, b, c. 

(8) For any elements a and b, we can find one and only one 

element x such that a + x = b. We call this element 
b — a. 

(9) There is a unique element 0 such that a + 0 = a for 
every element a. 

(10) For any elements a and b, provided only that a is not 0, 

there is one and only one element x such that ax = b. 

We call this element b/a. 


A Concrete Approa ch to Abstract Algebra 

(11) There is a unique element 1 such that for every a, a - 1 = a. 

The element 1 is not the same as the element 0. 

(12) a-b = 0 only if a = 0 or b = 0. 

(Students sometimes ask, “Ought we not include as 
an axiom that <2-0 = 0 for all a?” This, however, can 
be proved from the axioms we already have. By axioms (9) 
and (11), 1 +0 = 1. Therefore, a- (1 + 0) = a- 1. By ax- 
iom (7), a- 1 + a- 0 = a- 1. By axiom (11), a + a-0 = a. 
This says that x = a- 0 satisfies the equation a + x = a. 
Axiom (8) shows that this equation has only one solu- 
tion. Axiom (9) states that this solution is * = 0. So 
a- 0 = 0. Note that axiom (12) is intended to be read 
in this sense: “If you know that ab is zero, you can deduce 
that either a is zero or b is zero.” The result we have 
just proved is the converse of this.) 

In all of these axioms it should be understood that by 
“element” we mean an element in the structure. For 
instance, suppose we are applying the tests to the nat- 
ural numbers 0, 1, 2, 3, • • • . Someone might say, “Test 
(10) is passed, because if you take any quotient like 
3 4 it does exist; it is 3/4.” But 3/4 is not an element 

in the set 0, 1, 2, 3, • • • . It is true that by bringing in new 
elements 1/2, 3/4, -1, -2, and so on, you can obtain a 
field, the field of rational numbers in which division and 
subtraction are always possible. When we say that a 
structure is a field, we mean that it already contains 
the answers to every subtraction and division question. 
A child that only knows the numbers 0, 1, 2, 3, • • • , can 
only answer the questions “Take 4 from 3,” “Divide 3 
by 4” by saying “You can’t take 4 from 3,” “4 doesn’t 
go into 3.” This indicates that the natural numbers do 
not form a field; they fail tests (5) and (10). 

I have not attempted to reduce the tests to the smallest 
possible number, as might be done in a study of axiomat- 
ics. For instance, it is quite easy to show that a structure 

Arithmetics and Polynomials 


that passes tests (7) through (77) also passes (72). Some 
of my tests could be cut down somewhat; part could be 
assumed and the remainder proved. My purpose at 
present is to explain what a field is, and to give a speedy 
way of recognizing one. 


Determine which of the following are fields, and show on the 
chart which tests each passes. (At this stage you may find it 
best to convince yourself that certain properties do or do not 
apply, without necessarily being able to provide formal proof.) 

1 . The even numbers, 0, 2, 4, 6, • • • . 

2. The even numbers, including negative numbers, 0, ±2, 
±4, • • • . 

3. The real numbers. 

4. The complex numbers, x + iy, where x, y are real. 

5. The complex numbers, p -f- iq, where p, q are rational. 

6. The complex numbers, m + in, where m, n are integers. 

7. All numbers of the form p + qy/ 2, where p, q are ra- 

8. All expressions a + bx, where a, b are real numbers. 

9. All polynomials in x with real coefficients. 

10. All functions P(x)/Q{x), where P(x) and Q(x) are poly- 
nomials with real coefficients. 

1 1 . Arithmetic modulo 2. 

12. Arithmetic modulo 3. 

13. Arithmetic modulo 4. 

Question for Investigation 

If n is a positive whole number, what condition must 
n satisfy if the arithmetic modulo n is to be a field? It is 


A Concrete Approach to Abstract Algebra 

fairly easy to find out experimentally what the condition 
is. It is also easy to show that the condition is necessary; 
that is, that arithmetic modulo n cannot be a field unless 
n has a certain property. It is harder to prove that this 
property is sufficient to ensure the arithmetic being a 

A field is so much like our ordinary arithmetic that we 
can work with its elements just as if they were ordinary 
numbers ; our usual habits lead us to correct results, and 
we feel quite at home. 

But some structures, as we have seen, are provided 
with operations that we can label + and • , but yet fall 
short of being fields. One or more of statements (7) 
through (72) proves false. 

There are still other structures in which we do not 
have two operations, but only one. For example, we 
might consider the structure in which every element 
was of the form “x dogs and y cats,” x and y of course 
being positive whole numbers, or zero. It would be nat- 
ural to have the operation + defined to correspond to the 
word “and.” 

If a is 3 dogs and 4 cats, 

b is 5 dogs and 6 cats, 

a + b is 8 dogs and 10 cats. 

But there is no obvious way of defining the operation • ; 
we can hardly say that dog times dog is a square dog. 
We shall later meet less frivolous examples in which 
only one operation, either + or • occurs. 

These are, of course, simpler structures than arith- 
metic, and logically it would be reasonable to start 
with them and work up to arithmetic and other struc- 
tures with two operations. However, it seemed wiser to 
start with the familiar subject of arithmetic, and only at 

Arithmetics and Polynomials 


this stage to indicate that it occupies a fairly lofty posi- 
tion in the family of all possible structures. 

One might go beyond arithmetic to study structures 
with three operations, +, •, and * say. Whether any- 
thing of mathematical interest or value would be found 
in this way, I do not know. 

Polynomials Over Any Field 

The examples in chapter 1 suggested very strongly 
that most of the properties of polynomials in ordinary 
algebra were also true when we were working with the 
arithmetic modulo 5. It should be possible to generalize 
from this, and to find properties true for any field F — 
that is to say, for any system obeying axioms (7) through 

When we write the quadratic ax 2 -f- bx + c, we may 
have in mind, for example, 

(i) a , b, c integers, 

(ii) a, b, c rational numbers, 

(iii) a, b, c real numbers, 

(iv) a, b, c complex numbers, 

(v) a, b, c 0 or 1 in the arithmetic modulo 2, 

(vi) a, b, c 0,1, 2, 3, or 4 in the arithmetic modulo 5. 

In case (i) we say that ax 2 + bx + c is a quadratic 
polynomial over the integers. Thus llx 2 — Ax + 3 is a 
polynomial over the integers. 

In case (ii) we speak of a polynomial over the field of 
rational numbers; for example, \x 2 — fx + 

In case (iii) we have a polynomial over the field of 
real numbers; for example x 2 + -kx — e. 

In case (iv) we have a polynomial over the field of 
complex numbers; for example, (1 + i)x 2 + (J — 0* + 
(3 + 4f). 


A Concrete Approach to Abstract Algebra 

In case (v) we have a quadratic over the arithmetic 
modulo 2; for example lx 2 + Oat + 1. 

In case (vi) we have a quadratic over the arithmetic 
modulo 5; say, 2x 2 — f— 3x -f- 1 . This does not look any 
different from a quadratic over the integers. Perhaps if 
we write ( 2x 2 + 3x + 1) = 2(x -J- l)(x + 3), the dis- 
tinction will become apparent. You could, if you like, use 
a symbolism we had earlier, and write Hat 2 -f- III* + I, 
where the Roman numerals emphasize that we are 
dealing with numbers modulo 5. 

The word “over” has always seemed to me a little 
queer in this connection. Perhaps it is used because the 
coefficients can range over the elements of the field F. 
Anyway, all that matters is its meaning. ax n + bx n ~ x 
+ • • • 4- kx + m is a polynomial over the field F, if a , 
b, • • • m are all elements of F. The idea is a simple 

The Scope of x 

The step we have just taken corresponds to the begin- 
ning of school algebra, a, b, • • • , k, m are numbers of the 
arithmetic (see examples (i) through (vi) of the previous 
subheading). The symbol a is something new. We have 
passed from 11, —4, 3 to llx 2 — 4x + 3. 

The best pupils are not deceived by the apparent 
newness. They say, “You can test whether a statement 
about x is true by seeing whether it holds for any num- 
ber.” The worst pupils do not look at it this way. They 
have no idea what x means, but they manage to pick up 
certain rules for working with x. 

Curiously enough, both points of view are significant 
for modern algebra. They lead to an important distinc- 
tion. There are certain expressions (in certain fields F) 
that are equal when x is replaced by any number of the 

Arithmetics and Polynomials 


field, but they are not equal in the sense of being the 
same expression. The best pupils will say they are equal; 
the worst pupils will say they are not. 

An example will make this clear. Suppose our field F 
consists of the numbers 0, 1 of arithmetic modulo 2. If 
we ask a dull pupil “Is x 2 -\- x equal to 0?” the pupil will 
say “No.” We ask, “Why?” The pupil answers, “Well, 
they are different, x 2 + x is x 2 + x, and 0 is 0. They are 
two different things.” 

If we ask a bright pupil, who thinks of algebra as gen- 
eralized arithmetic, “Is x 2 + x equal to 0 in the arith- 
metic modplo 2?” this pupil will answer, “Let me see. 
If * was 0, x 2 -\- x would be 0. If x was 1, x 2 + x would 
be 1 + 1) which is 0. 0 and 1 are the only numbers in 
the field F. Yes; x 2 + x is always the same number as 0.” 

Actually, we have to regard both answers as correct. 
They are in effect answers to two different questions; 
they correspond to two different interpretations of equal. 
We shall need both of these ideas, and some agreed way 
of expressing them. 

If /(x) and g(x) are two algebraic expressions which, 
when simplified in accordance with the rules of an 
algebra, lead to one and the same polynomial ax" + 
bx n ~ l 4- • • • 4~ kx 4" m, we say that /(x) and g(x) are 
formally equal. 

If in /(x) and g(x), when we replace the symbol x by 
any element of the field F, the resulting values are the 
same, we say /(x) equals ^(x) for every x in the field F. 

Thus, for modulo 2 arithmetic, x 2 4" * and 0 are not 
formally equal, but they are equal for every x in the field. 

In ordinary algebra, it is not necessary to make this 
distinction. There is a well-known theorem that if /(x) 
and g(x) are two polynomials equal for all rational num- 
bers (or even for all integers), then /(x) and g{x) are 
formally equal. 


A Concrete Approach to Abstract Algebra 

In a field with a finite number of elements, this dis- 
tinction is bound to arise. For instance, arithmetic 
modulo 3 contains only the numbers 0, 1, 2. Evidently 
x(x — l)(x — 2) is zero for x = 0, for x = 1, and for 
x = 2. So x(x — l)(x — 2) = 0 for every x in the field. 
But if you multiply this expression out, you get an answer 
which is not Ox 3 -j- Ox 2 + Ox -f- 0. So x(x — l)(x — 2) is 
not formally equal to zero. 

Question: Multiply out x(x — l)(x — 2) modulo 3. 
Find and multiply out the corresponding expression for 
the arithmetic modulo 5. 

The fact just noted can be important. For instance, 
the equation x 2 = 2 has no solution in the field F con- 
sisting of 0, 1, 2 modulo 3. We might want to bring in 
a new sign, \^2, and extend our arithmetic just as we do 
with ordinary numbers. Now x(x — l)(x — 2) is not zero 
for x = \^2. If we are going to extend our number 
system in this way, it is the dull child’s answer and not 
the bright one’s that is helpful ! 

There are in fact three roles that x can play: (i) It may 
stand for any element of the field F. (ii) It may stand for 
certain elements outside the field F. (iii) It may not stand 
for anything at all — it may be just a mark on a calculat- 
ing machine. 

Ordinary high school algebra is a sufficient example 
of (i), where it is understood that any statement about 
x holds for every number of arithmetic. 

As an example of (ii) we might consider, say, %x 2 — 
-Jx — 5. This is a polynomial over the rationals. But we 
can consider the result of putting x = V / 2orx = 7 rin 
this expression. Neither number is rational. 

A more striking example can be found from calculus. 
(This is simply an example. Any student unfamiliar 
with calculus can omit it, as it is not needed for later 

Arithmetics and Polynomials 


developments.) Let f(t ) stand for any function of t that 
is capable of being differentiated as often as we wish. Let 
D stand for the operation d/dt so that 

W) = /'(*) 

DJ(t) = f"(t) 

D 3 f(t) = and so on. 

By an expression such as ( D 2 +2 D 3 )/(?), we shall 

understand /"(0 + 2/'(0 -j- 3f{t). (This example is in- 
tended to convey what we understand by addition of the 
operations D 2 , 2D, and 3.) 

Multiplication of operations means that the operations 
are to be applied successively. Thus, if I want to apply 
{D + 2) • (D 3) to f(t), I begin by finding (. D 4" 3 )f{t). 
Let this be u(t). I then apply the operation (D + 2) to 
u(t), and get ( D -j- 2 )u(t), that is, u!{t ) + 2 u(t), where 
u(t) = (£)-}- 3 )f(t) = f{t ) -fi 3 ft). If I substitute this 
value of u(t ) in u'{t) + 2 u(t) I get 

/"(0 + 3 f(t) + 2 fit) + 6/(0 = f’{t) + 5 fit) + 6/(0. 

To summarize what we have done: Applying the op- 
eration D 4- 3 to ft) gives f{t) 4- 3ft). Applying the 
operation D 4- 2 to the result above gives f'{t) 4~ 5/'(0 
4~ 6/(0« This last expression is what we understand by 
(D 4~ 2) • (D 4~ 3) acting on/(0- 

But the result above could be written ( D 2 4~ 3D 4“ 6) 
ft). That is: the operation (D 4~ 2) • (D 4~ 3) applied 
to any function has the same effect as D 2 4“ 3D 4- 6 
applied to the same function. We naturally call these 
two operations equal. In symbols 

(D 4- 2)-(D 4- 3) = D 2 4- 5D f 6. 

But this result has a familiar look. It is exactly what 
we should get if we forgot all about D standing for d/dt 
and simply applied the rules of elementary algebra. 

Thus, in certain circumstances, we are entitled to* 


A Concrete Approach to Abstract Algebra 

substitute in x 2 + 5x + 6 the value x = d/dt , which is 
not a number at all. This fact is made use of in teaching 
engineering students how to solve certain types of differ- 
ential equations. 

Finally, we may — as we did toward the end of the 
last example — say “let us forget all about the meaning 
of x, and simply remember than x may be manipulated 
according to certain rules.” This is the fully abstract 
approach. It corresponds to possibility (iii) above. 

The kind of theorem that can be obtained by the ab- 
stract approach is shown by the following important 
result. If you have a calculating machine with certain 
signs marked on it (the numbers or elements of F ) and 
also two operations, + and • , for which the twelve 
field axioms hold, then you can build a calculating 
machine that will have all those signs and also the signs 
x, x 2 , x 3 , • • • . On this enlarged machine, addition and 
multiplication will still be possible; the commutative, 
associative, and distributive laws will still hold. 

In other words, if you have any field F, and you in- 
troduce a new symbol x without inquiring at all into the 
meaning of x, and you perform calculations with poly- 
nomials in x (the coefficients being elements of F) by 
means of the ordinary rules of algebra, then you will not 
arrive at any contradiction. I have not yet proved this 
to be so. I have only indicated the kind of theorem that 
can be arrived at by the purely abstract approach. 

What do we actually use when we are making calcu- 
lations with polynomials? Suppose, for instance, we are 
multiplying (ax 2 -f- bx + c)(dx T e). If we proceed in 
great detail, we write 

(ax 2 T bx + c)(dx + e) 

— (ax 2 -f- bx -f- c)dx + (ax 2 + bx -f- c)e 
= ax 2 -dx 4" bx'dx -f- C'dx + ax 2 • e -f- bx • e + ce. 

Arithmetics and Polynomials 


So far we have used the associative law for addition, 
and distributive laws. These are not covered by axioms 
(7) through (72), which refer only to elements of the 
field F. But we have here also the symbol x, to which no 
meaning attaches, and which, accordingly, we cannot 
regard as being an element of F. 

To deal with a term like ax 2 • dx we have to use associ- 
ative and commutative laws for multiplication, and we 

(ax 2 )(dx) — a(x 2 -dx ) 

= a(dx-x 2 ) 

= a{d- xx 2 ) 

= a(d-x z ) 

= ( ad)x 3 . 

ad, of course, is an element of F . 

It would therefore be sufficient if we said, “Given a 
field F and a symbol x, we assume that the associative, 
commutative, and distributive laws hold not only for the 
elements of F, but also when the elements are combined 
in any way with x and its powers.” 

This would certainly serve as a very useful axiom and 
would allow us to establish the usual procedures for cal- 
culations with polynomials. But by what right could we 
bring in such an additional axiom? How do we know 
that such an element x can be joined on to any field? 
Might not a contradiction be produced by bringing in 
this extra axiom? Or at any rate, some extra information 
about F? There might be things we could prove about 
the elements of F themselves (that is, with no mention 
of x) with the help of this extra axiom, that we could 
not prove without it. 

Now in fact, with any field F, you can study polyno- 
mials over that field. In so doing you do not place any 
restrictions on F nor do you bring in any contradictions. 


A Concrete Approach to Abstract Algebra 

This is going to be something of a foundation stone for 
our future work, and we do not want to have any doubts 
about it. 

We want to show that if we have a calculating machine 
for an arithmetic 0, 1, • • • , we can always produce a 
calculating machine for an algebra 0,1, • • • , at. How are 
we going to get this new symbol x joined on to the exist- 
ing symbols 0, 1, • • • ? 

How could we show on a calculating machine the 
quadratic 2x 2 -j- 3x 4- 4? A quadratic is specified by 
three numbers — the coefficient of x 2 , the coefficient of x, 
and the constant term. A machine for displaying quad- 
ratics might have the following form. 

Constant Coefficient Coefficient 

term of x of x 2 

© ® © 

Suppose the machine had to add quadratics. It might 
appear as follows. 

First quadratic 




Second quadratic 








It is very noticeable here that we have no reference at 
all to x. On the outside of the calculating machine there 
might be painted (for the benefit of inexperienced op- 
erators) the words “Constant term; coefficient of x; co- 
efficient of x 2 .” But if the paint wore off, the machine 
would still work just as well. 

From the point of view of the designer of calculating 
machines, then, the quadratic a + bx + cx 2 is simply 
defined by the three numbers (a, b, c ). The quadratic 
a + j3x -j- yx 2 is something defined by the three num- 
bers (a, 7). The sum of the quadratics is something 

defined by the three terms (a -\- a, b (3, c 7). 

Arithmetics and Polynomials 


What about the product? 

(a + bx + cx 2 )(a + fix + yx 2 ) — aa + (a(3 + ba)x 

4" ( a 7 + b(i + ca)x 2 + (by -j- c0)x 3 -j- cyx 4 . 

The product is thus something defined by the sequence 
of numbers 

aa, afi + ba, ay b( 3 -f~ ca, by -J- c/3, ry. 

Our example above, with the calculating machine 
showing two quadratics, could thus be extended as 

First quadratic 



Second quadratic 









The machine so far seems designed only for multiply- 
ing quadratics. We should like it to deal with polynomials 
of any reasonable degree. A machine that would multi- 
ply polynomials, provided the answer was not above the 
seventh degree, would have the following appearance 
when set for the two quadratics above. 

First polynomial 








Second polynomial 


























You will notice how completely x has disappeared 
from the scheme. We think, of course, of the columns as 
corresponding to 1 , x, x 2 , x 3 , • • • , but this thought is, so 
far, not embodied in the machine. 

I do not know if you feel worried by a sense of the 
machine being incomplete. It would be unable to give 
the correct answer for the product of two polynomials of 


A Concrete Approach to Abstract Algebra 

the fifth degree, and it would not even be able to state 
the problem of multiplying two polynomials of the ninth 
degree. Only a machine with an infinite number of col- 
umns would be free from this kind of limitation, and one 
is naturally hesitant to talk about such a thing. 

We could get round this difficulty with a machine like 
an electronic calculator, where instructions go in on 
tape and a printed answer comes out. 

The problem would go in something like this. 

First polynomial: 4, 3, 2, finish. 

Second polynomial: 5, 7, 1, finish. 

Operation: multiply. 

The answer would come out as 

Answer: 20, 43, 35, 17, 2, finish. 

If the first polynomial had (m 1 ) numbers in it, and 

the second one (n -J- 1) numbers, the answer would 
contain (m + n 4- 1) numbers. 

Infinity does not trouble the construction of this ma- 
chine, and I think we could have easy consciences in 
talking about it. And yet, in a way, an infinity of zeros is 
implied, even in a finite expression like 4 + 3x + 2x 2 . 
If you ask me “What is the coefficient of x 3 here?” I 
answer “Zero.” If you ask me, “What is the coefficient 
of x 4 ” I answer “Zero.” The same for x 1,000 ’ 000 and 
^1, 000,000, ooo^ 

Thus, if I write “4, 3, 2, 0, 0, 0, 0, • • • , on forever,” 
I am not doing much different from making the finite 
statement, “4, 3, 2, finish.” 

Accordingly, I shall feel free to talk about 4, 3, 2, 0, 
0, 0, • • • , an infinite sequence, in the belief that this is 
not a vicious use of infinity. 

The operation that the machine does when it multi- 
plies two polynomials can be put in a more explicit form. 

Arithmetics and Polynomials 


If you look at the number ay -J- bfi + ca that occurs in 
the product of 

a b c 
a (3 y 

you may observe a pattern in it. This pattern can be felt 
in the muscles, if you put a forefinger on the Latin letter 
and a thumb on the Greek letter of the rows above, for 
each term, as you read ay T b(3 4~ ca - Your finger will 
move forward through a, i, c as your thumb moves back- 
ward through a, (3, y. 

A pattern of this kind appears most clearly if suffix 
notation is used. The product of 

pO pi p2 
qo qi qi 

contains the following numbers 

poqi 4- piqo, 
poq2 4" piqi 4~ p2q2, 

piq2 + p2qi , 


Question: What do you notice about the suffixes here? 
If we multiply 

po pi p2 * * * Pm 

qo qi ?2 * ’ * ' ’ ' qn j 

how can the resulting numbers be expressed most com- 

Your solution to 5 this question will, I am sure, be equiv- 
alent to the following. 

Let the product be ko, k\, • • • , k m+n . Then 

ko = poqo, 
ki = poqi + piqo, 

ki — poq 2 4" piqi 4“ P%qo, and so on. 


A Concrete Approach to Abstract Algebra 

In h the suffixes add up to zero, in ki they add up to 1, 
in k 2 to 2, and so on. In k r we expect the indices to add 
up to r. If p s occurs in k r , its partner must be q r _ s . We 
are thus led to write 

.<? — r 

kr ~ ^ ^ psC/r—s- 

s = 0 

It is understood, of course, that for s > m, p„ = 0 and 
for t > n, q t = 0. This corresponds to our remark that 
in the first row p m is followed by endless zeros, and sim- 
ilarly for q n in the second row. 

We have now arrived at a fair specification of a poly- 
nomial machine. A polynomial machine contains col- 
umns into which we can enter numbers of the field F. 
By “a polynomial of degree m” we understand that the 
machine is set so as to show “po, pi, pi, • • • , p m finish” 
or, equivalently, “p 0 , pi, pi, • • • , p m , 0, 0, 0, • • • ,” the 
zeros continuing forever. The input to the machine con- 
tains spaces for two polynomials, the first and second poly- 
nomials, say (po, pi, pi, • • •) and (q Q , q h q 2 , ■ ■ •)• The 
output of the machine shows two polynomials, namely 
the sum and product of the input polynomials. The sum 
is the polynomial (jo, ji, ji, • • • ), where j r = p r -\- q r . The 
product is the polynomial (k 0 , ki, k 2 , • • •), where 

s = r 

kr = ^ ] psqr— 8- 

s = 0 

The sum and product are defined for this new ma- 
chine. We cannot immediately identify sum on the new 
machine with + on the old, nor product on the new 
with • on the old. No doubt there is some relationship, 
but until we have established what it is, we had better 
use new signs for the new machine. I will use S for sum, 
P for product. 

The old machine is a calculating machine for the 

Arithmetics and Polynomials 


arithmetic of the field F. It adds and multiplies individ- 
ual numbers. 

The new machine deals with sequences such as (4, 3, 2, 
0, 0, 0, • • •) and (5, 7, 1, 0, 0, 0, • • •). It gives results 
such as 

(4, 3, 2, 0, 0, 0, • • •) S (5, 7,1,0, 0, 0, ■ ■ •) 

= (9, 10, 3, 0, 0, 0, ..•) 


(4, 3, 2,0,0, 0,-)P(5, 7,1,0, 0,0, •••) 

= (20, 43, 35, 17, 2, 0, 0, 0, ■ • 

We have noted above the rules by which the answers 
to ( ) S ( ) and ( ) P ( ) are obtained. 

It is important to note that nothing is stated about the 
new machine except that it carries out these rules cor- 
rectly. Nothing else whatever is assumed about the op- 
erations S and P. We know that the operations S and P 
can be carried out, because they depend simply on the 
operations of arithmetic in the field F. The old machine 
does arithmetic; the new machine will contain one or 
more replicas of the old machine, for doing the necessary 

The two machines, the arithmetic machine (the old 
machine) and the polynomial machine (the new ma- 
chine), are now set up. Their nature is determined. We 
cannot introduce any more assumptions. We can only 
observe the machines and see what they do. We have no 
more control over events. The machines must speak for 

We can however introduce certain abbreviations. It 
will be convenient to have a short name for the sequence 
(1, 0, 0, 0, • • •)• We shall call it 1. 1 is not the number 1. 
It is not a number at all; it is a sequence. In the same 
way 4 is a convenient abbreviation for the sequence 
(4,0,0, 0, •••)■ 


A Concrete Approach to Abstract Algebra 

The sequence (0, 1, 0, 0, 0, • • •) will also receive a 
name. If you think of how we originally arrived at these 
sequences, you will see what part this sequence is destined 
to play. It is the sequence we should use to represent the 
algebraic expression x. At last x is coming into the 

We accordingly define x as standing for the sequence 
(0,1,0, 0,0, •••). 

In case you think anything mysterious is involved in 
saying that 1 is a name for (1, 0, 0, 0, • • •) and * is a 
name for (0, 1, 0, 0, 0, • • •)> it may help you to see how 
simply these abbreviations can be embodied in the 
machine. All it means is that the operator has keys on 
which are written 1, 4, x, etc. If the operator presses 1, 
automatically the sequence (1, 0, 0, 0, • • •) appears. If 
the operator presses 4, then (4, 0, 0, 0, • • •) appears. If 
the operator presses the key marked x , then (0, 1 , 0, 
0, 0, • • •) will appear. 

Suppose, for instance, that in setting up the first poly- 
nomial the operator simply presses the key marked x, 
and does the same for the second polynomial. The ma- 
chine will carry out the operations S and P and we shall 
see the following. 

First polynomial, x, (0, 1, 0, 0, 0, • • •) 

Second polynomial, x, (0, 1, 0, 0, 0, • • •) 

Sum (0, 2, 0, 0, 0, • • • ) 

Product (0, 0, 1, 0, 0, • • •) 

You should check for yourself that the rules, by which 
the machine works, do lead to the results shown above. 

Thus (0, 0, 1 , 0, 0, • • • ) has turned up as the result of 
multiplying X and x. It seems natural to provide a key 
marked x 2 that will automatically set up (0, 0, 1, 0, 0, 
• • •). We shall then be able to note the result, which 
the machine gives us as 


Arithmetics and Polynomials 

(0, 1,0, 0, 0, •••)P(0, 1,0, 0, 0, •••) 

= ( 0 , 0 , 1 , 0 , 0 , 0 , • • •), 

in the shorter form, 

X P X = X 2 . 

Note that x 2 is a single sign. It is something marked on 
a particular key. When you press this key, you set 
(0, 0, 1, 0, 0, 0, •••)•! am not asserting that x 2 is the 
square of x or anything like that. You must remember 
that this is supposed to be the beginning of algebra. We 
have to pretend that you have never seen x 2 before; you 
know nothing of the implications of x 2 except what you 
can find out by operating the machine. 

Now we feel tempted to rush ahead and say, “I see 
how things are working out. (4, 3, 2, 0, 0, 0, • • •) is going 
to get the label 4 + 3x + 2x 2 , or in the new notation 
4S3PxS2P a: 2 .” Now that is true enough in the long 
run, but we have to be careful not to be carried away by 
the familiar notation and make unjustified assumptions. 
4 + 3x + 2x 2 has no meaning in a formal system until 
the associative law for addition has been established; for 
otherwise 4 + (3x + 2x 2 ) and (4 + 3x) + 2x 2 might have 
different values, and we do not know which is intended. 
Also we think of 4 + (3x -j- 2x 2 ) and (3* + 2x 2 ) + 4 
and 4 + (2x 2 + 3x) and (2x 2 + 3x) + 4 as all meaning 
the same thing. But this is not so, unless the commuta- 
tive law has been established. Even the use of 2x 2 nor- 
mally implies that (2*)* = 2(xx), that is, the associative 
law for multiplication. Different forms of bracketing give 
different results in a nonassociative system, x(x(xx)), 
((xx)x)x, and (xx)(xx) may have quite different values. 
You can see this by considering, in ordinary algebra, 
the symbol 



A Concrete Approach to Abstract Algebra 

“To the,” as used in “a to the is a nonassociative 

operation. Without some convention or bracketing, x x * X 
could have several meanings. 

(i) We might begin at the bottom and work up. Our 

calculation proceeds thus: ( x x ) x = x( x ^), and (x x ^) X 

= x(* 3 ). 

(ii) We might bracket thus (**)'*'• This equals 
**(**) = *(**+ 1 ). 

(iii) Finally we might work from the top down. This 

is the meaning normally attached to x xxX since both (i) 
and (ii) lead to answers that could be written more 

Working from the top down, for example, with 2 2^ 

would give 22 = 4, 2(2^) = 2^ = 16, and finally 2 2^ 
= 216 . 

Thus bracketing is needed to give a definite meaning to 
“x to the x to the x to the x,” 
and we must show good reason for it if we are going to use 
“x times x times x times x” 
without any brackets. 

We must in fact establish the five basic laws of algebra, 
which are usually written 

(I) u + v = v 4" u 
(II) u-v = vu 

(III) u 4- (v -fi w) — (u + v) + w 

(IV) u-(vw ) = ( uv)w 

(V) u(v + w) = ( uv ) + ( uw ). 

We have to establish these laws for polynomials, (se- 
quences) u, v, w. We have agreed to use and • for 

Arithmetics and Polynomials 


operations on the old calculating machine (the arith- 
metic of the field F). On the new calculating machine 
(for polynomials over F ), we have the operations S and P, 
and we must write the laws corresponding to (I) through 
(V) with these symbols thus 

(I') u S v = v S u 
(II') u P v = v P u 

(III') u S (v S w) = (u S v) S w 

(IV') u P (v P w) = (u P v) P w 
(V') u P (v S w) = (u P v) S (u P w). 

At least, we must do this if we want to be consistent. But 
the laws in the form first given, with -j- and • , look much 
more familiar and can be read at a glance. We can be 
forgiven if we sneak a look at them, to remind us of the 
part that laws (I') through (V') play. 

We are of course forced to use different signs for ad- 
dition and multiplication on the new machine; other- 
wise we should find ourselves assuming things in algebra 
that in fact we had only proved for arithmetic. Later, 
of course, when everything has been shown to work 
properly, we shall abandon P and S and replace them 
by • and -j- for future use. 

Now we are not in much doubt that laws (I') through 
(V') will work for polynomials, because, if you look back, 
you will see that we defined the operations S and P so 
that they gave a formal statement of what we do anyway 
in school algebra. 

There is, however, one point to watch. While, in illus- 
trations, I have been using the numbers of ordinary 
arithmetic — for example, 4, 3, 2, 0, 0, — the whole object 
of our present work is to show that we can build an 
algebra on any arithmetic. By an arithmetic here I mean 
a field F, a set of symbols obeying axioms (7) through 
(72). Now it might be that, in formulating these axioms, 


A Concrete Approach to Abstract Algebra 

we have overlooked some important property. To make 
sure that this is not so, we have to prove that our algebra 
works properly, without appealing to anything except 
axioms (7) through (72). If we can carry this proof 
through, then we know that axioms (7) through (72) 
not only express interesting properties of a number sys- 
tem, but that they contain all the conditions an arith- 
metic must satisfy if an algebra of polynomials is to be 
based on it. 

It is for this reason that some rather cautious formal 
work is now in order. We cannot say, “It’s a waste of 
time to prove such results in detail. Anyone can see that 
the results are true, and only drudgery is needed to carry 
the checking through.” The value of this formal work is 
not that it makes us more sure of the correctness of ordi- 
nary algebra but that it opens a whole new world to us. 
It assures us that if we have a system of elements, an 
arithmetic, we have only to verify that it satisfies the 
twelve axioms of a field, and then we shall be able to 
make algebraic calculations, on the basis of this arithme- 
tic, without learning any new habits. 

We now proceed to prove laws (I') through (V'), for 
polynomials over any field F. (See the comments on 
page 3 of the introduction before reading farther.) 
theorem (i'). If u and v are polynomials over F, then 
u S v = v S u. 

Proof. Let u be the sequence ( a 0 , a h a 2 , • * • ) and v the 
sequence ( b 0 , b h b 2 , • * •). It is understood here that is 
an element of F, and that for i > m, ai = 0. In the same 
way, b i is an element of F, and if i > n, b t = 0. 

Let u S v be ( e 0 , ei, e 2 , • • • ) and nSwbe (<7 0 , d\, d 2 , • • • ) . 

The definition of sum was given earlier. To find u 
S v, we take p r = a r , q r = b T in this definition. We find 
e r — a r ~f- b r for all r. To find v S w, we put p r — b r> 
q r = a r in this definition. 

Arithmetics and Polynomials 


We find d r = b r + a r for all r. But a r and b T are ele- 
ments of F. By axiom (3), a r -f- b r = b r + a T . 
e r = d r for all r. 

Therefore the sequences (<? 0 , e h e 2 , • • • ) and (d 0 , d h d 2 , 
• • •) are identical, u S v = v S u. Q.E.D. 

Although we have not mentioned it in this proof, 
axiom (7) is really involved here. Axiom (7) states that 
a + b is defined for all elements a, b of F. Without this 
axiom, the definition of S would be meaningless. 

In the same way, the definition of the operation P 
uses equations such as k 2 = p 0 q 2 + Mi + Mo- This 
equation has a meaning only in virtue of: axiom (2), 
which allows us to speak of the products p 0 q 2 , piqi, p 2 qo‘, 
axiom (7), which allows us to add these; and axiom (5), 
which allows us to omit brackets in the addition. 
theorem (n / ). If u and v are polynomials over F, then 
«P V = V P u. 

Proof: Let u P v be (/„, fi, f 2 , • • •) and v P u be 
(go, gi, g 2 , • • •)• We use the same symbols, (a 0 , a h a 2 , • • •) 
for u and (b 0 , b h b 2 , • ■ •) for v that were used in proving 
the preceding theorem. 

In the definition of a product, let p { = a { and q { — b { 
for all i. Hence, writing the summation in full, we 

fr — Mr + 0.\b r -\ -j~ a 2 b r —2 +•••-{- Mo- 
Similarly, on putting pi = b { and q t = a { , we obtain 
g r — b 0 a r -f- Mr-i + b 2 a r - 2 + • • • + Mo- 

By axiom ( 3 ), Mr + Mr-i = Mr-i + Mr- Using this 
axiom repeatedly, we can get Mr as the last term in the 
expression for g r instead of the first. We can similarly 
work the term M r -i into the last place but one, by use of 
the same axiom. We can continue to rearrange the terms 
of g r until the order has been completely reversed. Hence 


A Concrete Approach to Abstract Algebra 

by repeated use of axiom (3), with axiom (5) in the back- 
ground to justify the omission of brackets, we find 

g r = b r ao + b r -\d\ + • • • + b\d r —\ -j- bod r . 

In each product, now use axiom (4). 

gr = d 0 b r + d\b r - 1 + • • * + d r —]b\ + d r bo = f r . 

Hence u P v = v P u. Q.E.D. 

THEOREM (ill') . U S (v S w) = (u S v) S W. 

Proof. Let u be («o, cl\, < 22 , • • •), v be (bo, bi, bi, • • •), 
and w be (t 0 , c\, Ci, •••)• 

u S (v S w) implies that we first combine v and w, and 
the result of this is combined with u. v S w is (bo + co, 
bi + ci, bi + d, • • • ) • Hence u S (v S w) is 

(ao + [bo + Co], a,\ + \b\ + c{\, d2 + [bi + C2], • • •)• 

We now calculate the quantity on the right-hand 
side of the equation. 

u S v is 

(do + bo, a\ + b\, d 2 + bi, •••)• 

(u S v) S w is 

([<zo + bo] + Co, \d\ + b\] + ci, [di + bi] + Ci, • • •). 

By axiom (5), a, + (b t + c l ) = (d { + b % ) -f a for all i, 
and hence the sequences are identical. 

u S (v S w) = (u S v) S w Q.E.D. 

THEOREM (iv') . U P (v P w) — (u P v) P W. 

Proof. We keep the symbols for u, v, w that were used 
in proving the preceding theorem. 
v P w is the sequence 

(boCo, boC\ -f- b\Co, boCi + b\C\ + biCo, 

boCz + b\Ci + biCi + b$Co, •••)• 

u P (v P w) will thus be the sequence (ho, hi, hi, h s , • • •), 

Arithmetics and Polynomials 


ho — a Q (boCo), 

hi — ao(boC\ + b\Co) + a\(boCo ), 

h = ao(boC 2 + bid + b 2 c 0 ) + di(boCi + bic 0 ) + d 2 (boCo), 
and so on. 

We arrive at (u P v) P w by seeing that (u P v ) is 

(dobo, dobi + aibo, dob 2 + aib\ + d 2 bo, 

dobz~\~ d\b 2 -f* d 2 bi + azbo, • * •)• 
So (u P v) P w is (ko, k h k 2 , • • •), where 

ko — (d 0 bo) Co, 

hi = (dobo)ci -f- (d 0 bi -f- dibo)co, 

k 2 = (dob^)c 2 + (dobi + dib 0 )ci + (d 0 b 2 + d X bi + d 2 b 0 )co, 
and so on. 

Now, in these particular cases, it is easy to multiply 
out the expressions for ho, hi, h 2 , and ko, k x , k 2 and see that 
equal expressions result, ho, h, h 2 can be multiplied out 
directly by appealing to axiom (7). We cannot apply 
axiom (7) directly to finding k x and k 2 , since axiom (7) 
deals with a(b + c). It says nothing about (b -f- c)d ! 
But this does not delay us for long. 

For (b + c)d = a(b + c), by axiom (4) 

— db + dc, by axiom (7) 

— bd + cd, by axiom (4). 

It is easy to deduce that 

(b + c + d)d = bd + cd + dd, 
since b + c + d = (b + c) + d. The result above has to 
be used twice. Similarly we prove 

(b + c + d + • • • + j)a = bd + cd + dd + • • • + ja. 

We can thus verify ho = ko, hi = k h h 2 = k 2 . This 

makes it plausible that h r = k r for all r, but of course 

it does not prove that it is so. It would require an eternity 
to work out all the individual results. To finish our proof 
we must find some way of describing h r and k r , from which 
it will become evident that h r = k T for all r. 


A Concrete Approach to Abstract Algebra 

If you look at fa you will notice that the suffixes of a, 
b, c add up to 2 in each term. In fact, fa contains all the 
terms a s b t c w for which .r + t + w = 2, and each combina- 
tion 5 , t, w occurs once only. The same is true of fa. 
(There is nothing surprising about this result. It pro- 
vides, in fact, a method of elementary algebra for mul- 
tiplying polynomials.) 

In general, we can show that both h r and k r are equal 
to 'La s b t c w over all j', t, w for which s + t w = r. 

The only difficulty in giving a proof of this result is a 
difficulty very characteristic of logical analysis. There is 
a stage when the facts to be proved are so evident that 
you do not know just how much commentary is called 
for. One faces this kind of difficulty, for instance, if 
asked to prove 4 + 5 = 9. If someone questions 4 + 5 
being 9, we wonder what he accepts that will serve as a 
basis for the proof. 

The same kind of difficulty arises here. It is pretty 
evident that h r does contain all the terms a s b t c w such 
that + t + w = r. The difficulty is to see just what one 
must say to justify it. The simplest course is to write, 
“Obviously it is so.” 

I think what we have to establish is the following. 

(i) If s + t + w — r, a s b t c w occurs in the expression 
for h T . 

(ii) It occurs once only. (Note that aibtfs is not the 
same as a 2 b 3 c\. Both of these will occur in fa.) 

(iii) No term occurs in h r other than the type men- 
tioned in (i) above. 

The same of course must be proved for k r . 

Conditions (i), (ii), and (iii) are easily verified. Let 
DPwbe (jo, ji, ; 2 , • • •)• Then, by the definition of P, 

ji = boCi + biCi-i + b 2 Ci- 2 + • • • + biCQ. 

Since u P (v P w) is (fa, fa, fa, • • •), we have 

Arithmetics and Polynomials 


hr = a 0 jr + ai/r- 1 + jr —2 + * * * + Orjo- 

We want to prove, for (i), that if .$• + f + a; = r, then 
a s b t c w occurs in h r . 

Now h r contains a s j r -s • As r — s = t + w, a s j r - s = 
cisjt+w jt+w contains b t c w . Hence the term a s b t c w does 
appear in h r . (It cannot cancel, since no negative terms 
are involved.) 

Next, for (ii), we must show that a s b t c w occurs once 
only. In h r , a s occurs only in a s j T - s * And j r - s , that is, 
jt+w, contains only one term involving b t , namely b t c w . 
Thus a s b t c w cannot occur more than once. 

Finally, we prove (iii). h r is the sum of expressions 
a s jr- s , and j r - s is the sum of terms b t c r - s -t ■ Hence every 
term appearing in h T is of the form a s b t c r -s-t , that is to 
say, of the form a s b t c w where r — s + t + w. 

We have now completed the proof that h r is as stated. 
A similar argument can be carried through for k r . Hence 
h r — k r for all r, and Theorem (IV') is proved. Q.E.D. 
THEOREM (v') . U P (v S w) = (u P v) S (u P w) . 

Fortunately the proof of this theorem is much shorter. 
It consists simply of a direct verification. 

v S w is the sequence ( b 0 + c 0 , bi + c\, 62 + ^2, ’ ’ ’)• 
u P {v S w) is then (y 0 , yi, yz, ’ • •), where 

Jr = Go {b r + Cr) + a\(b r -l + Cr- 1) + - * - + G r (^o + A))- 

On the right-hand side, u P v is (/o, fi, f 2, • • •)> where 
f r = aob r + a\b r -\ -(-•••+ a r bo. 

«Pa)is (zoj z\, Z2, • • •), where 

Z r — Go Cr “t“ CL\C r —\ OrCt). 

* The trouble with this kind of analysis is that you start asking 
yourself, “How do I know that a s only occurs once? Ought I to 
prove that it does?” Every analysis must stop somewhere. We 
may as well be firm. We have no doubt that the statement just 
made is true. We stop here. Future generations can get their 
Ph.D.’s in mathematical philosophy by analyzing this statement 
if they want to. 

56 A Concrete Approach to Abstract Algebra 

The theorem requires us to show that y r = f r + z T for 
all r. But 

fr + z r = (aob r + af) r -\ + • • • + a r bo ) 

+ (flo C r + d\C r —\ + * * * + d r cf) 

— (do br + aoCr) + (d\b r -\ + fliCr-l) + • • * 

+ (a r bo + a r co) by repeated use of axioms 
(3) and (5), 

— do (b r + C r ) + d\(b r -\ + C r - 1) + * * * 

+ a r (b 0 + co) on applying axiom (7) to each 

= yr- 

Hence the theorem is proved. Q.E.D. 

In these proofs it will have become apparent that 
many of the steps we take in algebraic calculations can 
be justified by appeal to the commutative, associative, 
and distributive laws. We began by assuming these for 
the arithmetic of the field F ; we have now shown, by 
establishing (I') to (V'), that these same principles also 
hold for polynomials over F. This allows us to relax our 
vigilance somewhat. We now know that it is quite justi- 
fied to assume for polynomials u, v, w that uv may be 
replaced by vu , that u(v + w) may be multiplied out as 
uv + uw } that uvw has a meaning without any need for 

We next proceed to put the theory into a more famil- 
iar form. We have already introduced the abbreviations, 

1 for the sequence (1, 0, 0, 0, • • •), 
and x for (0, 1, 0, 0, • • •), 

and x 2 for (0, 0, 1,0, • • •). 

definition. x n stands for the sequence with p n — 1, all 
other pi zero. 

Note here that p r = 1 refers to the element 1 of the 
field F. The n that occurs in p n and x", however, is an 
ordinary counting number of our everyday arithmetic. 

Arithmetics and Polynomials 


For instance, in the arithmetic modulo 5 with the ele- 
ments O, I, II, III, IV, we still consider sequences such 
as (II, I, III, O, IV, II, I, O, O, • • •) with p 0 = II, 
pi = I, p 2 = HI, p% = 0,p k = IV, p 5 = II, p 6 = I, and 
we can consider II - II - II * II * II * II * II = II 7 . Thus, even 
though the arithmetic modulo 5 only contains the five 
distinct elements O, I, II, III, IV, yet the ordinary 
numbers, including the numbers 5, 6, 7, • • • , cannot be 
kept entirely out of it. This might cause complications 
in the Pacific Island experiment (p. 18). 

THEOREM. X m P X n — X m+n . 

Proof. Let x m be (p 0 , pi, p 2 , • • •) where p m = 1, all 
other elements being zero. 

Let x n be (q 0 , qi, q 2 , • • •) where q n = 1, all other ele- 
ments being zero. 

If x m P x n is (£o, k\, k% , • • •), then 


k r = ^ ] p&qr—f 

s = 0 

The term p s q r - s will be zero if either p s or q r - s is zero. 
p s is non-zero only for s — m, and q r - s is non-zero only 
if r — s = n. Hence a non-zero term occurs on the right- 
hand side only if s = m and r — s = n; that is, only if 
r = m + n. Hence k r is zero except for r = m + n, and 
km+n = 1 • 

But k m + n =1, all other k r zero is the sequence for 

X m+n 

x m P x n = x m+n . Q.E.D. 

Theorem. If a is any element of F, and a denotes the se- 
quence (a, 0, 0, 0, • • •) then aP / is the sequence (0, 0, • • • , 
0, a, 0, 0, • • • ) ; that is, ax n is the sequence defined by p n = a, 
all other pi = 0. 

Proof. Calculate a P x n . Verification is immediate. 
theorem. Every polynomial can be written in the form 
p 0 S (pi P x) S (p 2 P x 2 ) S • • • S (p n P x"). 


A Concrete Approach to Abstract Algebra 

The meaning of this theorem can be appreciated most 
easily if, for a moment, we replace S, the sum sign, by 
the usual +, and P, the product sign, by • . The expres- 
sion above then becomes 

pO + pi'* + P2-* 2 + * • • + P»-*"- 

Proof. Since pa is (po, 0, 0, 0, 0, • • •), while by the pre- 
vious theorem 

px P * is (0, pi, 0, 0, 0, • • •), 
p 2 P * 2 is (0, 0, p 2 , 0,0, • • •), 

and so on, it is evident that (po, pi, pi, po, • • • , p n , 0, 0, 0, 
• • • ) can be obtained by summing the terms above. 


Thus, apart from the fact that the unfamiliar S and 
P are playing the parts of + and • , we have now reached 
a stage where polynomials can be written down in the 
ordinary way, and added and multiplied; we know that 
the commutative, associative, and distributive laws, (I') 
through (V'), hold — that is to say, that the processes 
behave as in ordinary algebra. 

There is, however, one remaining point. We seem now 
to have two distinct subjects. In terms of our machinery, 
we have one machine for adding and multiplying the 
elements of F — the numbers of the arithmetic. This is 
the old machine. The new machine deals only with se- 
quences, with polynomials. The two machines seem quite 
distinct; no cables pass from one to the other. This seems 
a little peculiar, for surely an arithmetic and its algebra 
should be connected? 

In our ordinary algebra we speak of a quadratic poly- 
nomial such as 4 + 3* -f- 2x 2 , a linear polynomial such 
as 4 + 3x, and a constant polynomial such as 4. Thus, one 
and the same sign, 4, is used for the number 4 and the 
constant polynomial 4. Nor does this cause any trouble. 

Arithmetics and Polynomials 


Most people in fact are hardly aware that any such dis- 
tinction can be made. Probably this distinction is in- 
volved in the trouble that some students find in drawing 
the graph of y = 4. They seem to be helped by the sug- 
gestion that they should draw the graph of y = 4 + 0 • x, 
which brings out the kinship of y = 4 to the family of 
lines y — mx -f- c, or reminds the student of the procedure 
for tabulating values and drawing the graph. The psy- 
chological effect of writing y = 4 as y = 4 — f- 0 • at seems 
to lie in emphasizing that 4 can be regarded as a par- 
ticular case of a polynomial. 

How does this overlapping of number behavior and 
polynomial behavior appear on our machines? Let a and 
b be any elements of F. On the arithmetic machine, we 
can find a + b and a-b. What will happen if, on the 
polynomial machine, we find a S b and a P b? 

a is {a, 0, 0, 0, • • •)• 
b is ( b , 0, 0, 0, * * •)■ 

The rules for S and P give 

a S b is {a -j- b, 0, 0, 0, • • •). 
a P b is ( ab , 0, 0, 0, • • •)• 


a S b = (a + b), 
a P b = (ab). 

For example, 

2 S 3 = 5, 

2 P 3 = 6. 

This means to say, we can do our arithmetic on the 
algebra machine. 

2 S 3 = 5 means that the polynomial 2 added to the 
polynomial 3 gives the polynomial 5; this runs exactly 
parallel to the relation between numbers, 2 + 3 = 5. 
This parallelism is an example of isomorphism. The struc- 


A Concrete Approach to Abstract Algebra 

ture, the pattern, of numbers is exactly the same as the 
structure, the pattern, of constant polynomials. 

Accordingly, it becomes something of a luxury to own 
both a polynomial machine and an arithmetic machine. 
Since we can do all our calculations on the polynomial 
machine, we can trade our arithmetic machine in. When- 
ever we want to do an arithmetical calculation about 
numbers a, b, c, we can set up the polynomials (a, 0, 0, 
• ' ( b , 0, 0, • * •), ( c , 0, 0, • • •) and infer the arithmet- 

ical result from operations on the polynomial machine. 

Getting rid of the arithmetic machine is welcome. We 
have so far had to use S, P, and +, • to avoid confusion 
between operations on polynomials (sequences) and op- 
erations on numbers. But now that the arithmetic ma- 
chine has gone, there is no longer a need for this distinc- 
tion. We can write the familiar + where S stood on the 
machine, and • for P. We can also drop the distinction 
between the number a and the constant polynomial a. 

In effect, the sequences ( p 0 , p h p 2 , • • •) and the opera- 
tions S and P are now relegated to the interior of the 
machine. On the outside of the machine we see (if ordi- 
nary numbers compose the field F) 0, 1, 2, 3, and so on; 
+ and • ; *, x 2 , x 3 , and so on. 

If we press 4 -f- 3-x + 2-x 2 , then, in some space inside 
the machine, the sequence (4, 3, 2, 0, 0, 0, • • •) will ap- 
pear. If we simply press 4, the sequence will be (4, 0, 0, 
0, • • •)• When the operation + is required, the machine 
carries out S in its interior; when • is required, the ma- 
chine carries out P. 

Note that * has not changed its meaning at all. When 
we press x, the machine sets up (0, 1, 0, 0, 0, • • •) in its 
interior. That is all we know about x. There is no re- 
quirement that * be an element of F, a number, or any- 
thing of that sort. 

Arithmetics and Polynomials 

Substituting a Value for x 


Have we perhaps gone too far? After all, we still want 
to say that x 2 — 3x + 2 = 0 if * = 1 or x — 2. What are 
we to do when we want to substitute x — 2 or x = k in 
a polynomial po + p\x + — + p n x n ? 

x m has been defined by the sequence (0, 0, 0, • • • ,0, 
1 , 0, 0, 0, • • • ) that it produces in the machine. However, 
the theorem we had earlier, which we now write x m -x n 
— x m+n gives us another way of regarding x m . 

We have seen earlier that x 2 — x-x. Hence 

x 3 = x 1+2 = x-x 2 = x-x-x, 
x 4 = x 1+3 = x-x 3 = x-x- x-x, 

and so on. The cubic polynomial 

a -f- b-x + c-x 2 + d-x 3 

could thus be written 

a -f- b-x + c-x-x + d-x-x-x. 

In what relation does this stand to the expression 

a + b-k + c-k-k + d-k-k-k , 

where k is an element of F? 

Let us think what happens in the interior of the ma- 
chine when these two things are calculated: 

a produces (a, 0, 0, 0, 0, 0, 0, • • •), 

b-x produces ( 0 , b, 0, 0, 0, 0, 0, • • •), 

c-x-x produces (0, 0, c, 0, 0, 0, 0, — ), 

d-x-x-x produces (0, 0, 0, d, 0, 0, 0, • • •). 

Summing a + b-x + c-x-x + d-x-x-x produces (a, b, c, 
d, 0, 0, 0, • • • ) as, of course, we could have foreseen. On 
the other hand, 


A Concrete Approach to Abstract Algebra 

a produces {a, 0, 0, 0, 0, 0, 0, • • •)> 
b-k produces {bk, 0, 0, 0, 0, 0, 0, ■ • •)> 
c-k-k produces ( ck 2 , 0, 0, 0, 0, 0, 0, • • •)> 
d-k-k-k produces ( dk 3 , 0, 0, 0, 0, 0, 0, • • •). 

Summing a + b-k + c-k-k + d-k-k-k produces 
(a + bk + ck 2 + dk\ 0, 0, 0, 0, 0, 0, • • •)• 

It will be convenient to use the abbreviation /(x) for 
a + b -x + c-x-x + d-x-x-x, and f{k) for a + b-k + 
c-k-k + d-k-k-k. 

These look very much alike, but they appear very 
differently on the machine. /(x) appears as ( a , b, c , d, 
0, 0, 0, • • •), a polynomial of degree three, specified by 
the four elements a, b, c, d of F. f(k) appears as ( e , 0, 0, 

0, 0, 0, - - •), where e — a + b-k + c-k-k + d-k-k-k is 

a single element of F. We can regard f(k) as a constant 
polynomial or, if we like, we can identify it with the ele- 
ment e of F; we agreed earlier to neglect the distinction 
between e, the constant polynomial ( e , 0, 0, 0, • • •) and 
the element e of F. 

In ordinary high school algebra we meet the idea of 
“substituting k for x.” On our calculating machine we 
accordingly must introduce a key marked, “substitute 
k for x.” 

If the calculating machine is carrying the sequence 
(a, b, c, d, 0,0,0, - - - ) which represents /(x), and we press 
the “substitute k for x” key, the machine will compute 
e = a + bk + ck 2 -f dk z — this is a single number, an 
element of F — and will set up the sequence ( e , 0, 0, 0, 
0 , 0 , •••). 

This explains and defines what we mean by “substi- 
tute k for x.” If on a machine you see a knob marked 
with an unfamiliar word, and you want to understand 
the meaning of this word, all you need do is turn the 
knob and observe what happens in the machine. When 

Arithmetics and Polynomials 


you are completely familiar with, what this knob dogs , 
then you understand the word written on it. There is 
nothing more to understand. 

I emphasize this because there is much discussion at 
present of the correct use of words. People are worried 
about whether they really understand what a word 
means. Some philosophers hold that successful thinking 
is only possible after all words have been defined exactly. 
It may be that some people actually do think by means 
of words. If so, they are a minority. Most people think 
by imagining, by picturing actual objects, actual proc- 
esses, actual happenings. We speak to each other in order 
to make sure that we are thinking of the same thing and 
picturing it in the same way. We clear up confusion in 
our own minds by thinking about something until we 
have a clear picture of it. 

The idea that a word or a symbol has one fixed, exact 
meaning is dangerous even in mathematics. Poincare, 
one of the greatest modern mathematicians, declared 
“Mathematics is the art of calling different things by 
the same name.” We have had an example of this 
already. We used + originally in arithmetic with 
3 -}- 4 = 7 . When we got to dealing with the arith- 
metic modulo 5 , we could have brought in a new sign 
of combination and written, perhaps, III © IV = II. 
But we found it more convenient to write 3 + 4 = 2. 
2 , 3 , 4 , and + all have new meanings here, but we find 
it convenient to use the same signs. When we were 
dealing with sequences, we used a new sign S for sum, 
3 for (3, 0, 0, 0, • * ■), and 4 for (4, 0, 0, 0, 0, • • •)• We 
were glad, though, when we stopped writing 3 S 4 = 7 
and went back to 3 + 4 = 7 for constant polynomials. 
So we have already used + in three different senses. 
When we use a sign, then, it does not imply that all its 
meanings are exactly the same, but that there is enough 

€4 A Concrete Approach to Abstract Algebra 

in common to make the family resemblance worth 

I should certainly not advise anyone teaching 
ninth-grade algebra to say that x is the sequence 
( 0 , 1 , 0 , 0 , 0 , 

Earlier we discussed the question of the polynomial 
x 2 + x over F, the field consisting of 0, 1 modulo 2. Was 
this polynomial equal to zero or not? We have two sets 
of facts: 

(i) The polynomial x 2 + x is the sequence (0, 1, 1, 0, 
0, 0, • • •). The constant polynomial 0 is the sequence 
(0, 0, 0, 0, 0, 0, • • • ) • These two sequences are distinct. 
In this sense, x 2 + x 5 ^ 0. 

(ii) If in x 2 + x we substitute 0 for x, we get 0 2 + 0, 
that is, 0. If we substitute 1 for x we get l 2 + 1, which 
is also 0. 0 and 1 are the only elements of F. Hence for 
every k of F, k 2 + k = 0. 

It is desirable that we should have some way of indi- 
cating whether we are posing questions, or making 
statements, in the sense of (i) or (ii). We might make 
some sort of convention about the use of letters; we 
might always use x for sense (i), and k for sense (ii). 
This might be inconvenient at times; it is surprising how 
soon the letters of the alphabet get used up in a mathe- 
matical discussion. 

It is better to have some terminology to indicate how 
a symbol is being used. When we write k 2 + k = 0 to 
mean that if any element of F is added to its square, the 
result is zero, we say that A: is <2 variable. When we write 
x 2 + x without caring what x may mean or whether it 
means anything at all, simply as a symbol obeying cer- 
tain formal rules, we say that x is an indeterminate. 

A professor of mathematics once made the following 
criticism of the remainder theorem. He said, “The re- 

Arithmetics and Polynomials 


mainder theorem states that when f(x) is divided by 
x — a, the remainder is f{a). But when x = a, x — a is 
zero. You cannot divide by zero. That means we cannot 
consider x being a when we are dividing by x — a. The 
proofs of the remainder theorem that depend on a being 
substituted for x are therefore false.” 

This criticism is entirely incorrect. The critic is clearly 
thinking of x as a variable , so that x is a number and 
x — a is also a number. Let us take an example of divi- 
sion and see how well or badly it works to regard x as 
a variable. 

Example. “When x 2 is divided by x — 3, the quotient 
is x + 3 and the remainder 9.” I am sure you will agree 
that this is a correct statement. Let us see how it looks 
when we take particular numbers for x. 

x = 5. “When 25 is divided by 2, the quotient is 8 
and the remainder 9.” 

x = 6. “When 36 is divided by 3, the quotient is 9 
and the remainder 9.” 

x = 7. “When 49 is divided by 4, the quotient is 10 
and the remainder 9.” 

There is a sense in which all of these are true. For 
instance, 25 = 2X8 + 9. Yet they represent very 
strange arithmetic. 

If, purely as a matter of arithmetic, we had been asked 
to divide 25 by 2, 36 by 3, and 49 by 4, we would have 
given the remainders as 1, 0, 1, respectively. We could 
never have arrived at the conclusion “The remainder is 
always 9” by arithmetical experiments. 

In fact, the sentence “When x 2 is divided by x — 3, 
the remainder is 9” is a statement of algebra that is not 
a generalization about arithmetic. The division in ques- 
tion is not arithmetical division: it is division according 
to the rules we learn for dealing with polynomials. 


A Concrete Approach to Abstract Algebra 

“Divide x 2 by x — 3” means exactly the same as 
“divide (0, 0, 1, 0, 0, 0, • • •) by (-3, 1, 0, 0, 0, • • •)” 
It does not make sense in any other interpretation. 

The thing we are dividing by here is ( — 3, 1, 0, 0, 
0, • • •), and this is not zero. Zero is (0, 0, 0, 0, • • •). 
There is no need to take precautions to ensure that the 
divisor does not become zero. ( — 3, 1, 0, 0, 0, • • •) just is 
not (0, 0, 0, 0, 0, • • •), and there are no circumstances in 
which it could become (0, 0, 0, •••). A polynomial can 
always be divided by x — 3 in the sense in which long 
division is done in algebra. 

The distinction between x regarded as a variable and 
x regarded as an indeterminate is therefore not an empty 

We do have to distinguish between the two statements 

Statement I. For the indeterminate x , the polynomial 
/(x) = 0. This means that /(x) is (0, 0, 0, • • *)• This 
statement is true, for instance, if /(x) = — x 2 + 1 + 
(x — l)(x + 1). When this /(x) is simplified by the rules 
of algebra, the result is zero. We could express it as 
(1, 0, -1, 0, 0, 0, • • •) S (-1, 1, 0, 0, • • •) P (1, 1, 0, 0, 
0, •••) = (0,0, 0,0, •••). 

Statement II. For all values of the variable x, the 
polynomial /(x) = 0. This means that, for every element 
k of the field F, we have f(k) = 0. 

What is the relationship between statement I and 
statement II? We have seen that statement II is true, 
with F the arithmetic modulo 2, for x 2 + x. But state- 
ment I is not true for x 2 + x, since x 2 + x is (0, 1, 1, 
0, 0, • • • ) and not (0, 0, 0, 0, 0, • • • ) . So statement I does 
not follow from statement II, at least not for every field 
F. (It does follow in elementary algebra, as we noted 

Does statement II follow from statement I? If state- 

Arithmetics and Polynomials 


ment I is true, f(x) is (0, 0, 0, 0, 0, 0, • • •)• Now (0, 0, 0, 
0, 0, 0, • • •) is a particular case of ( a , b, c, d, 0, 0, • • •) 
with a = 0, b = 0, c = 0, d = 0. I have quite arbitrar- 
ily chosen to regard (0, 0, 0, 0, 0, • • • ) as a cubic, in order 
to make clear that we are not involved in a discussion of 
the infinite series 0 + 0k + Ok 2 + 0£ 3 + 0£ 4 + • • • . A 
polynomial is essentially a finite sequence. In the lan- 
guage of page 42, it may be “4, 3, 2, finish.” “Finish” 
means that we have reached the stage where we have 
zeros only. (0, 0, 0, 0, 0, • • •) could be written “finish”; 
with it we finish before we start! This makes the inter- 
pretation of the procedure for substituting k slightly 
obscure. If, however, we regard (0, 0, 0, 0, 0, • • •) as 
“0, 0, 0, 0, finish,” which we are entitled to do, we can 
apply the rule without any difficulties of interpretation. 
The explanation of “Substitute k for x” given on page 61 
leads to the value e = a + bk + ck 2 + dk 3 = 0 + 0& 
+ 0 k 2 + Ok 3 = 0, whatever element of F the quantity 
k may be. Hence, statement II follows from statement I. 

There is an important point to notice. The explanation 
of “substitute k for x” assumed that f(x) had been 
obtained in the standard form, po + pix + p 2 x 2 + • • • 
+ p n x n . Substituting k for x gave po + p\k + p 2 k 2 + • • • 
+ pnk n . 

If we were given /(x), say in the form 

f(x) = (ao + a\x + a 2 x 2 )(bo + b\x + b 2 x 2 ), 
the definition instructs us to multiply out /(x) as 
^ij ^ 2 , dj fij ' ' P (b 0 , b\, b 2 , 0, 0, 0, ’) — (aobo, 

Qob\ H - a.\bo, CLobi -f- a\b\ + a^bo, a\b 2 H - Qibi, a 2 b 2 , 0, 0, 0, 


and then to press the key “Substitute k for x,” giving 
f(k) = aobo -t~ (aobi -j- a^bo) k + (aob 2 + a\b\ + a^bo) k 2 + 
(a\b 2 -f- a 2 bi) k 3 -f- a 2 p 2 )&. 

But you naturally ask, “Gould we not use the simpler 
form (a 0 + a\k + a 2 k 2 )(b 0 + b\k + b 2 k 2 )?” The sugges- 


A Concrete Approach to Abstract Algebra 

tion is that multiplying out and then substituting could 
be replaced by substituting and then multiplying out. 
For instance, the two procedures for substituting 3 for x 
in (* — 1)(* + 1) would be 
(i) (x — l)(x + 1) = x 2 — 1. 

Substituting 3 for x gives 3 2 — 1 = 8. 

(ii) Substituting 3 for x in (x — 1) gives 2. 

Substituting 3 for x in (x + 1) gives 4. 2*4 = 8. 
Procedure (i) we know to be correct, for it merely carries 
out the instructions of the definition. Procedure (ii) we 
observe, in this example, gives the same answer as pro- 
cedure (i). 

Now, of course, it is not surprising that this should be 
so. If you will look back to page 34, you will see that we 
obtained our rules for constructing the polynomial ma- 
chine by considering what happens when we add and 
multiply in elementary algebra. The process of our 
thought has been something like this. 

(I) We learn arithmetic. 

(II) We notice in arithmetic various facts, such as 
3 X 4 = 4 X 3 and 3 + 4 = 4 + 3. We satisfy our- 
selves that these facts are particular examples of general 
properties of numbers, expressed by axioms (7) through 
(72). The numbers of arithmetic form a field. 

(III) The processes of elementary algebra can be 
justified by the field axioms. For instance, if a, b, c, d are 
elements of a field F, we can show that 

(a + bk) -f- (c + dk) = (a + c) + (b + d)k 


(a + bk) (c + dk) = ac + (ad + bc)k + bdk 2 

for any element k of the field F. 

(IV) We extract from (III) the abstract pattern 

(a, b, 0,0, • • •) S (c, d, 0, 0, • • •) 

= (<s + c, b + d, 0, 0, • • •), 


Arithmetics and Polynomials 

{a, b, 0,0, •••) P (c, d , 0, 0, • • •) 

= ( ac , ad + bd , 0, 0, • • •), 

and specify, quite abstractly, our polynomial machine 
for combining sequences by the operations S and P. 

(V) By introducing the sign * for the sequence 
(0, 1, 0, 0, • • •)> an d slipping back into writing + for S 
and • for P, we are able to write the equations of (IV) 
in the form 

(a + bx) + (c + dx ) = (a + c) + (b + d)x, 

(i a + bx)(c + dx) = ac + (be + ad)x + bdx 2 , 

which look very much like the equations of (III). What 
we have gained is that x is no longer restricted to being 
a number, an element of the field F. 

(VI) We verify that the sequences of (IV), and con- 
sequently their expressions in the symbolism of (V), obey 
the commutative, associative, and distributive laws, and 
hence can be worked with in exactly the same way as 
the expressions of elementary algebra. 

(VII) The operations of (IV) and (V) have been 
based on the patterns of (III). This means that, although 
the symbol x does not have to be interpreted as an element 
of the field F, yet it can be so interpreted. If, in any true 
statement of (V), * is everywhere replaced by k, an ele- 
ment of the field F ', we obtain a true statement in the 
language of (III). 

The assertion (VII) may be illustrated by the follow- 
ing two theorems. 

theorem. If p(x), q(x), f(x) are polynomials in the inde- 
terminate x over a field F, and p(x) + q(x) = f(x), then for 
any element k of F, p(k) + q(k ) = f(k ). 
theorem. If p(x), q(x), g(x) are polynomials in the inde- 
terminate x over a field F, and p(x) • q(x) = f(x), then for any 
element k of F, p(k) • q(k) = f(k ). 

By combining these theorems, we can obtain results 


A Concrete Approach to Abstract Algebra 

such as this: if 

/(*) = P(x) q(x ) + r(x) 
for an indeterminate x, then 

f{k) = p(k ) q(k) + r(k) 
for any element k of F. 


1 . Write out the proof of the remainder theorem. When any 
equation is written, make it plain whether the symbols are 
indeterminates, variables, or fixed elements of a field F. State 
explicitly any theorems used in the course of the proof. 

2. ax 2 + bx + c is a quadratic over a field F in an indeter- 
minate x. Explain what is meant by saying that the quadratic 
has at most two roots. Prove that this statement is correct 
whatever the field F. 

Chapter 3 

Finite Arithmetics 

Earlier we considered the arithmetic of Even and 
Odd. This was a particular example of an arithmetic 
modulo n, namely the arithmetic modulo 2. When we 
were discussing Even and Odd, we accepted many state- 
ments such as “Even plus Odd equals Odd” as agreeing 
with our experience of arithmetic. We did not analyze 
or prove these results. To prove them is, of course, simply 
a matter of elementary algebra. We now look at these 
proofs as a first step toward proving the properties of the 
arithmetic modulo n. 

“Even plus Odd equals Odd” is a short way of saying, 
“When any even number is added to any odd number, 
the result is an odd number.” To prove the truth of this 
statement we must translate into algebraic symbolism 
the expressions “even number” and “odd number.” 

If our symbols a, b,c, • • • , x,y, z stand for integers, 
the numbers 2a, 2b, 2c, ••• , are even; the numbers 2 a + 1, 
2b + 1 , 2c + 1 , • • • , are odd. Any even number can be 
expressed as 2a. Any odd number can be expressed as 
26 + 1. When these are added, we get 2a + 2b + 1 = 
2 (a + b) + 1, hence an odd number; and we have 
proved the statement. 

There is no difficulty in justifying the other results of 
the addition table modulo 2. 


A Concrete Approach to Abstract Algebra 

The multiplication table is justified by the following 
four results: 

(2a) • (2b) = 2(2 ab), 

(2a) ‘(2b + 1) = 2-a(2b + 1), 

(2a + l)-2 b = 2-b(2a + 1), 

(2a + 1) • (2b + 1) = 2(2 ab + a + b) + 1. 

Apart from the particular results obtained, such as 
“Odd times Odd equals Odd,” this procedure shows that 
the question “When an odd number is multiplied by an 
odd number, is the result Odd or Even?” has a definite 
answer. We do not need to ask, “Which odd numbers 
are you thinking of?” 

Compare the situation if we divide numbers up into 
perfect squares and other numbers. One cannot answer 
“yes” or “no” to the question, “When a square number 
is added to a square number, is the result square?” 0 + 1 , 
9 + 16, 25 + 144 are examples of results that are 
squares; 1+4, 1 + 9, t 4 + 9 are examples of results 
that are not. 

Our procedure then, in forming the arithmetic modulo 
2 (and the other modular arithmetics), has been to divide 
the integers up into classes in such a way that, if we know 
to which classes x and y belong, we know to which classes 
x + y and xy belong. 

The arithmetic modulo 2 is easy to talk about because 
we have two generally accepted words, Even and Odd, 
to describe the classes. The language does not contain 
words to express that a number is divisible by 3, or leaves 
remainder 1 or remainder 2 on division by 3. We accord- 
ingly have to invent symbols such as 

O to label any number of the form 3a, 

I to label any number of the form 3a + 1, 

II to label any number of the form 3 a + 2. 

Finite Arithmetics 


Similarly, we used the symbols O, I, II, III, IV for the 
arithmetic modulo 5. 

For example, to prove that III -IV = II modulo 5, 
means to prove that (5a + 3) (5b + 4) is of the form 
5 c + 2, which is very easy to do. We thus show (i) that 
when a number x from class III is multiplied by a num- 
ber y from class IV the product xy lies in a definite class, 
independent of the particular numbers x and y chosen; 
(ii) that the class is in fact the class II. 

Of these results, (ii) is, so to speak, arithmetical. It 
holds for arithmetic modulo 5, but not necessarily for 
other arithmetics; for instance, in modulo 7 arithmetic, 
III IV = V. 

Result (i), however, has no reference to particular 
numbers, and generalizes immediately to give theorem 1 . 
theorem 1. If, for any whole number n, the integers are 
divided into classes, so that class P contains all numbers of the 
form na + p, class Q contains all numbers of the form na -f- q, 
and so on; then, given only that the numbers x andy belong to the 
classes X and Y, the classes to which x + y and xy belong are 

We can restate this result as follows: if x\ and x 2 are in 
the same class, and y\ and y 2 are in the same class, then 
xi + y\ and x 2 + y 2 are in the same class, and xiyi and 
x 2 y 2 are in the same class. 

Saying that xi and x 2 are in the same class is equivalent 
to saying that x\ — x 2 is a multiple of n. 

Thus we are given = nc + x 2 , yi = nd + yz- 

Hence *i + yi = n(c + d) + x 2 + y 2 and (x\ + yi) — 
(x 2 + yf) is a multiple of n. 

Also xiyi = n 2 cd + ncy 2 + ndx 2 + x 2 y 2 so xiyi — x 2 y 2 is 
a multiple of n. 

The theorem is proved. Q.E.D. 

We have now shown that axioms (7) and (2) hold for 
arithmetic modulo n. We have defined X + Y and X-Y; 


A Concrete Approach to Abstract Algebra 

namely, if x is any number of the class X, and y any 
number of the class Y, then X + Y means the class to 
which x y belongs, and X - Y the class to which xy 
belongs. Theorem 1 assures us that this definition is 

Axioms (3), ( 4 ), (5), (6), (7) are now almost self- 

Axiom (3) requires X + Y = Y X. Now T + ^ is 
the class containing y + x, while X + Y is the class con- 
taining x + y- But we know, for any numbers x, y, that 
y + x = x + y. Call this number k. Then X -f- Y is the 
class containing k, and Y + X is the class containing k. 
Hence X + Y and Y + X are the same class. Q.E.D. 

Axioms ( 4 ), (5), (6), (7) follow in the same way. If Z 
is the class containing z, axiom (7), for instance, re- 
quires X-(Y + Z) = (X-Y) + (X-Z). We show that 
X-(Y + Z) is the class containing x(y + z), and (X-Y) 
+ (X-Z) is the class containing xy + xz. As the numbers 
are the same, the classes are the same. 

Axiom (8), on subtraction, asserts that, for given 
classes P, Q (containing the numbers p , q) the equation 

P + X= Q 

has one and only one solution. It evidently has a solution. 
For let .S' be the class containing the number (q — p). 
Then P + S is the class containing/) +(<?—/>), that is, 
P + S contains q. .*. P + S = Q. S is a solution. 

Suppose the class T, containing the number t, is also 
a solution. P + T is the class containing p + t. If 
P + T — Q, this means that/) -f- t belongs to Q. .'./) + t 
= q + na for some whole number a. t = (q — p) + na. 
Hence t belongs to the class S. .\ The class to which t 
belongs is S. That is, T is S. 

Axiom (8) is thus proved to hold. 

Axiom (9) asserts that there is a class playing the role 

Finite Arithmetics 


of zero. Let O be the class containing the number 0. 
P + O is the class containing the number p + 0. As 
p 0 = p, P + O = P. Hence O plays the role of zero. 
By axiom (8), there can only be one such class. Other- 
wise the subtraction P — P would have more than one 

We may as well deal with axiom (77) now. It should 
be evident that the class I, containing the number 1, 
plays the role of unity. 

Axioms (77) and (72) do not hold for all modular 
arithmetics. For instance, in arithmetic modulo 6, the 
equation 11 -X — III has no solution, so division is not 
possible. Also II -III = O; axiom (72) does not hold in 
this arithmetic. 

The class O contains all numbers of the form na + 0, 
that is, all numbers na , all multiples of n. Thus “ XY = O 
only if X — O or Y — O” is the same requirement as 
“ry is a multiple of n only if x is a multiple of n or y is a 
multiple of n This property is characteristic of prime n. 

Hence axiom (72) holds if and only if n is prime. 

Does axiom (10) hold when n is prime? That is to say, 
given classes P, Q, where P is not O, can we find a class 
X such that P-X = Q? 

How do we go about finding such an X in practice? 
For example, in the arithmetic modulo 5, how should we 
solve II -X = III? We should have to look through the 
II times table until we found the answer III. Here is the 
two times table. 

no = o 

II -II = IV 
II -III = I 

If you look at the right-hand side, you will see that all 
the classes O, I, II, III, IV occur there. That is to say, 


A Concrete Approach to Abstract Algebra 

we may make II -X equal to any class we like, by choos- 
ing X suitably. In arithmetic modulo 5 then, division by 
II is always possible. 

This suggests a way of studying the arithmetic modulo 
n. This contains n classes, corresponding to the numbers 
0, 1, 2, • • • , (n — 1). We take some class P, and let X 
run through the n classes. If the resulting PX are all dif- 
ferent, they must form a complete set of classes. By choos- 
ing X suitably, we can make PX — Q for any assigned 
Q. Division by P is then possible. 

Can we show then that, when P is not O, all the P X 
are different? We suppose, of course, that n is prime; it 
is easily seen that for n composite the result is not true. 

The proof is extremely simple. It almost leaps to the 
eye as soon as we state the question in algebraic symbols. 

Suppose the PX are not all distinct. Then X = U 
and X = V give the same result; that is PU = PV. But 
(in view of the axioms already proved) this means 
P{U — V) = O. But P is not O, by our assumptions. 
U — Vis, not O, since U and V are not equal. Axiom (72) 
has already been established. Hence P(U — V ) is not O, 
since neither factor is O. We thus arrive at a contradic- 
tion if we assume the PX not distinct. Hence the PX are 

Accordingly, when n is prime, division is possible, for 
divisors other than O. We have thus theorem 2. 
theorem 2. The arithmetic modulo n, where n is prime, is 
,a field. 

For this arithmetic satisfies axioms (7)-(72). 

You may have noticed that we proved axiom (70) 
from axiom (72) and the remaining axioms. It is evident 
that there is a close connection between axioms (70) and 
(72). Since we proved axiom (70) from axiom (72), have 
we perhaps not been extravagant in including axiom (70) 
at all? Could we not simply assume (72) and deduce (70)? 

Finite Arithmetics 


For a finite field — that is, a field with a finite number 
of elements — we could do so. But consider a very familiar 
structure, the integers. For the integers, axioms (7) 
through (9), (77), (72) hold. All the answers in the two 
times table are distinct: 

-6, -4, -2, 0, 2, 4, 6, ••• . 

But this does not prove that every number occurs as an 
answer in the two times table. In fact, 2x = 3 has no 
solution in integers. 

Thus for the integers, axiom (72) holds, but (70) is 
false. It is thus impossible to deduce axiom (10) from (72), 
the other axioms being given, without the additional 
information that the structure contains only a finite 
number of elements. And most of our algebra, of course, 
deals with structures having an infinity of elements. For 
such a structure, axiom (70) is a stronger assumption 
than (72). 

Question: If axioms (7) through (77) are given, does 

Axiom (72) follow? 

An Alternative Approach to Division 

Our proof of axiom (10) has two limitations. First of 
all, a proof has two functions. One is to establish the 
particular result. This, of course, our proof does. The 
other function is to throw light on similar problems, to- 
suggest analogies. Our proof holds only for finite struc- 
tures. For infinite structures it is, if anything, misleading. 

Second, our proof shows that PX = Q has a solution, 
but it does not offer us any convenient way of calculat- 
ing that solution. We are assured that, if we try the n 
possible values of X, we shall find a solution among them. 
Our work will not be wasted; but if n is at all large, we 
may have a lot of work to do . 


A Concrete Approach to Abstract Algebra 

An alternative approach is possible, which both en- 
ables us to calculate X and sheds light on related prob- 
lems. It leads us to a theorem that does not appear very 
extraordinary, but which in fact is of great importance. 
This alternative approach is through a process that you 
will find in the older arithmetic textbooks for finding the 
highest common factor, or H.G.F. (The H.G.F. is also 
known as the greatest common divisor, or G.G.D.) Sup- 
pose the H.G.F. of 481 and 689 is required. The work is 
set out as follows: 













The two numbers 481 and 689 are first written down. 
We divide 481 into 689. It goes once and leaves remain- 
der 208. We divide this remainder into 481. It goes 
twice, and leaves 65. This remainder 65 is then divided 
into 208. It goes three times and leaves 13. Finally 13 
divides 65. It goes exactly five times. The remainder 0 
has now been reached, and the process terminates. The 
last number reached before zero, namely 13 here, is the 
H.C.F. As 481 = 37 X 13 and 689 = 53 X 13, this re- 
sult is correct for this example. 

Of course, with different initial numbers, the process 
may run to different lengths. We will justify the process 
in the particular case in which it follows the pattern 
above. It is easy to see that the ideas of this proof apply 
equally well in the general case. 

Suppose, then, that we begin with two numbers a, b 
and obtain the following scheme. 

Finite Arithmetics 
















This is equivalent to the following equations. 

b — ma — c (1) 

a — nc = d (2) 

c — pd = e (3) 

d — qe = 0 (4) 

We want to prove that e is the H.C.F. of a and b; that 
is, we want to show that a and b are multiples of e , and 
that no number larger than e is a factor of both a and b. 

From equation (4), d is a multiple of e. 

From equation (3), c — pd + e, so c is a multiple of e. 

From equation (2), a = nc + d. As c and d are both 
multiples of e, so is a. 

From equation (1), b — ma + c. As a and c are mul- 
tiples of e , so is b. This establishes that a and b are mul- 
tiples of e. Now suppose that some number k is a factor 
of both a and b. 

Equation (1) shows that k is also a factor of c. As k is a 
factor of a and c, equation (2) shows that k is a factor of d. 
Then equation (3) shows that A; is a factor of e. 

Hence any common factor k of a and b is a factor of e. 
It is thus impossible that k should be larger than e. Hence 
e is the H.C.F. of a and b. Q.E.D. 

We can draw a further consequence from these equa- 
tions. In equation (3) we can substitute for d from equa- 
tion (2). This gives 

e = c — p(a — nc) = c( 1 + pn ) — pa. 

In this equation we can substitute for c from equation (1). 


A Concrete Approach to Abstract Algebra 

e — (1 + pn)(b — ma) — pa 
— (1 + pn)b — a(m + pmn + p)- 

Thus e has been expressed in the form ax + by where x 
and y are integers. It does not disturb us that x happens 
to be negative. 

The letter h is frequently used for the H.G.F. Using 
this notation, we have the important theorem 3. 
theorem 3. If h is the H.C.F. of the integers a , b, there 
exist integers x, y such that 

h — ax + by. 

For example, the numbers 31 and 40 have H.C.F. 1. 
Thus it must be possible to find whole numbers x, y 
for which 

31* + 40 y = 1. 

To find these by trial and error would not be too easy. 
If we carry out the H.C.F. process and apply the argu- 
ment used above, we find that x — — 9, y = 7 is a solu- 
tion. (Other solutions also exist.) 


1. Find integers x, y to satisfy (i) 3x + 5y = 1, (ii) 8x + 
13 y = 1, (iii) 17x + 12 \y = 1. 

2. Could there be integers x, y for which Ax + 6y = 1? 

3. What is the necessary and sufficient condition that k must 
satisfy if ax + by — k is to have a solution in integers x, y? 
a, b are given integers. 

We are now in a position to deal with the question of 
division in the arithmetic modulo n, with n prime. 

If, as before, P and Q stand for the classes to which 
given numbers p and q belong, solving PX = Q is equiv- 

Finite Arithmetics 


alent to finding a number x such that px is in class Q. 
This means that, for some a , px = na + q. Conversely, 
if we can find numbers x, a to satisfy this last equation, 
then PX — Q for the class X containing x. 

Now n is prime, p is not a multiple of n, for that would 
mean P = O. The only factors of n are 1 and n itself. 
Since n is not a factor of p , the H.C.F. of n and p must 
be 1. Accordingly, by Theorem 3, we can find integers 
u, v for which pu + nv = 1 . 

Multiply this equation by q. 



puq + nvq = q. 


x = uq, a — —vq. 

px — na = q. 

Hence PX = Q as required. Note that, if u belongs to 
the class U, 

PU = I. 

In the modulo n, arithmetic, U is I/P, the reciprocal 
of P. Theorem 3 has in effect enabled us to show that 
every class P has a reciprocal I/P. If we multiply this 
reciprocal by Q, we arrive at Q/P, the desired quotient. 


1. In the arithmetic modulo 17, find the reciprocals of 7 
and 11. 

2. In the arithmetic modulo 257, find the reciprocals of 43 
and 16. 

3. In the arithmetic modulo 11, simplify the fractions 2/7 
and 5/6. 


A Concrete Approach to Abstract Algebra 

The calculations above are simply intended to help 
you to fix the content of theorem 3 in your mind. Need- 
less to say, it is the light this theorem throws on general 
theory that is important, not its use in calculations such 
as question 2 above. We rarely need to perform calcula- 
tions of this type. 

Note that the proof above makes no mention of the 
finiteness of the structure. The line of thought here 
followed proves fruitful in connection with infinite fields, 
as we shall see. 

Chapter 4 

An Analogy Between Integers 
and Polynomials 

There are many resemblances between the integers 
' ' ’ > — 5, — 4, — 3, — 2, — 1, 0, 1, 2, 3, • • • , and the poly- 
nomials a, bx + c, dx 2 + ex + /, • • • , where a , b, c, d, e, 
f, • • • , stand for rational numbers. Actually, I do not 
have to insist on the coefficients a, b, c, , being ra- 
tional numbers. They could be chosen from any field 
without harming the analogy. I specify the rational field 
in order that we may have something definite and 

The resemblance will become apparent if you run 
through axioms (7)-(72), and see which of them are 
true for our polynomials. If you add two polynomials, 
the result is a polynomial. If you multiply two polyno- 
mials, the result is a polynomial. Thus axioms (7) and 
(2) hold. Axioms ( 3 ) through (7), the familiar laws of 
algebra, certainly hold. Axiom (5) holds; subtraction can 
be done with polynomials. Axiom (9) holds; when you 
subtract a polynomial from itself you get zero. Axiom 
(70) does not hold. x/(x ,+ 1) is not a polynomial. Axiom 
(77) does hold; 1 is a polynomial. Axiom (72) also holds; 
if a product of polynomials is zero, one factor must be 


A Concrete Approach to Abstract Algebra 

zero. Here, of course, “a polynomial is zero” means that 
the polynomial is 0 + 0* + Ox 2 . Axiom (72) amounts to 
saying that, when you start out to multiply, say, x — 1 
and x + 1 , neither of which is zero, you can be sure that 
the product will not be 0 + Ox + Ox 2 . The product is, 
of course, x 2 — 1. You are speaking in quite a different 
sense if you say, “But x 2 — 1 is zero if x is —1 or 1.” We 
went into this distinction in chapter 2. 

The structure formed by the polynomials thus satisfies 
all the field axioms except (70). If you will look back to 
the chart in chapter 2, you will see that the integers did 
exactly the same. 

Question: Do the integers and the polynomials con- 

stitute isomorphic structures? 


In a field, division can be carried through without 
remainder. In the structures we are now considering, 
that cannot be done owing to the absence of axiom (70). 
We can however carry out division in an amended sense. 

44 divided by 9 gives 4 with remainder 8. 
x 2 divided by x — 1 gives x + 1 with remainder 1 . 

How shall we describe what we do when we divide 44 
by 9? We might say that we consider the multiples of 9, 
and see which of them comes nearest to 44. This suggests 
that 45 does, and that we might say 44 divided by 9 gives 
5 with remainder — 1 . This might indeed give a simpler 
and neater theory of division. We will, for the moment, 
keep both possibilities in mind. The multiples of 9 are 
the numbers 9 n. How near 9n is to 44 is estimated by 
considering 44 — 9n. We might make a table. 

Analogy Between Integers and Polynomials 



44 — 9 n 













As we go away from n = 4 and n = 5, the numbers in 
the second column get numerically larger. +8 and — 1, 
we may say, are the simplest numbers in the second 
column. It is a matter of opinion whether +8 or —1 is 
the simpler. In grade school arithmetic, 8 is definitely 
simpler, because it would take us so long to explain what 
we meant by —1. To someone familiar with the integers, 
however, —1 might well seem simpler than +8, since 
| — 1| is smaller than |8|. 

Whichever view we adopt, we can say that the quo- 
tient is that number for which 44 — 9 n is simplest. The 
remainder is the corresponding value of 44 — 9 n. 

By either convention, the remainder is always simpler 
than the divisor. By the grade school convention, the 
remainder 8 is a simpler number than the divisor 9. By 
the other convention, since | — 1| < |9|, —1 is simpler 
than 9. 

This kind of division, then, implies that we have some 
way of deciding whether one element is simpler than 

We know from experience that we can do long division 
with polynomials until we reach a remainder of lower 
degree than the divisor. Accordingly, we are led to make 
the convention that a polynomial /(*) is simpler than a 


A Concrete Approach to Abstract Algebra 

polynomial g(x) if fix) is of lower degree than g(x). We 
might expect to go further, and explain how we compare 
polynomials of equal degree, but for division this is not 
necessary. If polynomial P(x) is to be divided by D{x), 
there is only one Q(x) that makes P(x ) Q(x)D(x) of 

the lowest possible degree. This Q(x) is called the quo- 
tient, and P(x) - Q(x)D(x) is then the remainder R(x). 

For example, x 2 - Q(x)(x - 1) is a constant only if 
Q(x) = x + 1. The remainder R(x) is 

* 2 - (x + 1) (x - 1), that is, 1. 

R(x), of course, is not always a constant. For instance, 
when x 3 H - 2 is divided by x 2 1, Q(x) — x makes 
(*3 2) — Q(x)(x 2 — 1) the linear polynomial x + 2, 

and no other choice of Q(x) will lead to any simpler 
result. Thus R(x) = x + 2. 

The ordinary process of long division is equivalent to 
a series of steps in each of which the highest power of x 
is removed. Thus, when dividing x 4 + x 3 + x 2 + x + 1 
by x — 1, we first subtract from x 4 + x 3 + x 2 + x + 1 
the quantity (x — l)x 3 , which leaves 2x 3 + x 2 + x + 1. 
From this we subtract (x - l)2x 2 and obtain the quad- 
ratic 3x 2 + x + 1. From this we subtract (x — l)3x and 
obtain 4x + 1. Finally we subtract (x — 1)4, which 
leaves 5. All these steps can be combined in the statement 

(x4 + x 3 + x 2 + x + 1) ~ (x - l)* 3 ~ (x ~ l)2x 2 

— (x — l)3x — (x — 1)4 = 5. 

In virtue of the distributive law (V'), this means that 

(x 4 + x 3 + x 2 + x + 1) 

— (x — l)(x 3 + 2x 2 + 3x + 4) = 5. 

Question: Give an interpretation of the steps in the 

usual arithmetical process of long division, when 12,345 
is divided by 38 . 

Analogy Between Integers and Polynomials 



With integers or polynomials, division usually leaves a 
remainder. It may however happen, as when 12 is di- 
vided by 3, or x 2 — 1 by x — 1, that the remainder is 
zero. In this case we say that the divisor is a factor. 3 is a 
factor of 12, x — 1 is a factor of x 2 — 1. d is a factor of p 
when, for some integer q, qd = p. D(x) is a factor of P(x) 
when, for some polynomial Q(x), Q(x)D(x) = P(x). 

The numbers 1 and —1 are factors of every integer. 
For any integer n, n — 1-n — ( — 1 )•( — «). This factor- 
ization is trivial. 

definition. An element that is a factor of every element of 
a structure is called a unit. 

What are the units in the system of polynomials? We 
can eliminate at once any polynomial of the first or 
higher degree. For such a polynomial, P(x ), is certainly 
not a factor of P(x) + 1 . There remain only the constant 
polynomials. Apart from zero, these are factors of every 
polynomial. For instance, 2 is a factor of x + 1 since 
x + 1 = 2(|x + ^). You will remember that at the out- 
set of this chapter we said we should accept as eligible 
any polynomial with rational coefficients. 

Since any constant non-zero A: is a factor of any poly- 
nomial, we also regard these factors as trivial. A teacher, 
wishing a class to see that x 6 — 1 could be factored as the 
difference of squares or as the difference of cubes, might 
ask, “Is there any way of starting to factor x 6 — 1 
other than (x 3 — l)(x 3 + 1)?” A pupil might suggest 
(2x 3 — 2)(|x 3 + J), and the teacher might say, “Well, 
that is really the same factorization, isn’t it?” “Really 
the same” means that only a trivial operation has been 

In the same way, a pupil who factors x 2 — 1 as 


A Concrete Approach to Abstract Algebra 

( 2x — 2) + |) has undoubtedly given a correct an- 

swer. One might point out that (x — 1)(* + 1) is a 
somewhat more convenient form of this answer. 

A prime element is one that has no factors except trivial 
ones. Thus the factorization ( — 1)*( — 3) does not pre- 
vent 3 from being a prime number. The factorization 
2(i* + |) does not prevent x + 1 from being a prime 

Note that we call an element prime when it has 
no factors within the structure. For example, 3 = 
(Vl - 2)(a/ 7 + 2) and x 2 - 2 = (x - V2)(x + V2). 
But V7 - 2 and V7 + 2 are not integers, and x — V2 
and x + V2 are not polynomials with rational coeffi- 
cients. 3 and x 2 — 2 are prime elements of the integers 
and of polynomials over rationals respectively. 

Highest Common Factor 

d is a common factor of a and b if a = pd, b — qd, 
with p and q elements of the structure. 

We now have to explain what we mean by the 
“highest” common factor. We are looking, of course, 
for an explanation that will hold, not only for the inte- 
gers, but for any structure reasonably like the integers. 

In chapter 3 we met a procedure for determining the 
H.G.F. of two given integers a, b. These integers were 
then thought of as being positive, but since the H.C.F. 
of 8 and 12 is the same as the H.C.F. of 8 and —12, 
this is no real restriction. This procedure led us to a 
number h with the following properties: 

(i) h is a common factor of a and b. 

(ii) h = au + bv for some u, v belonging to the 

(iii) any common factor of a and b is a factor of h. 

In property (ii) we say “belonging to the structure” 

Analogy Between Integers and Polynomials 


instead of “which are integers” so that we shall not have 
to reword the statements when we proceed to consider 
structures other than the integers. 

It is evident that property (iii) is a direct consequence 
of property (ii). Property (iii) is stated here because it 
links on to our experiences in elementary arithmetic, 
which statement (ii) does not. 

We shall consider only structures in which a procedure 
analogous to the H.C.F. calculation of chapter 3 can be 
carried through. We suppose this structure to satisfy the 
field axioms, with the exception of (10). The proof in 
chapter 3 that h has properties (i)— (iii) depends only on 
these axioms. 

Question: Check this last statement. 

Any element h having properties (i)— (iii) will be called 
a H.C.F. Later we shall discuss what is involved in 
speaking of the H.C.F. 

The H.C.F. process depended on repeated divisions. 
The field axioms, with (10) removed, are insufficient to 
guarantee division. As we saw in our analysis of division, 
we need (1) a definition of “u is simpler than v”, (2) the 
property that, given any elements a, b, it is possible to 
find an element q such that r = a — qb is simpler than b. 

We suppose our structure has a definition (1) and 
satisfies requirement (2). This is still not quite enough. 
The H.C.F. process of chapter 3, in the example there 
considered, gave the sequence of numbers 689, 481, 
208, 65, 13, 0. These numbers get steadily smaller 
(“simpler”). This is no accident. For instance, 65 arises 
as remainder for a division by 208. So it is bound to be 
simpler than 208. The same holds for the other numbers 
in the sequence. The H.C.F., 13, is the number that 
occurs immediately before zero. Thus, an essential fea- 
ture of the process is that it must terminate. Otherwise, 

90 A Concrete Approach to Abstract Algebra 

there is no sense in speaking of the element just before 
the end. 

We therefore require of our structure (3) that every 
sequence of elements, u\, u 2, M3, • • • , for which u T+ 1 is 
simpler than u r , terminates. It is impossible to have an 
unending sequence of elements, each simpler than its 

Here, of course, it is understood that the sequence has 
a first term u\. For the integers one could have an infinite 
sequence, • • • , 5, 4, 3, 2, 1, 0, without a beginning. The 
H.C.F. process automatically provides initial elements 
a, b. The sequence is bound to have a beginning. We de- 
mand that it shall also have an end. 

In any structure, then, satisfying the field axioms with 
the exception of (10), and requirements (1) through (3) 
just listed, we can carry through the H.C.F. process, and 
find an element h with properties (i) through (iii). 

In particular, the H.C.F. procedure can be carried 
out for polynomials over the rationals. We verify re- 
quirements (1), (2), (3), as follows. 

A polynomial corresponds to a sequence (a, b, c, • • • , 
k, 0, 0, 0, • • • ) . The complexity of a polynomial is meas- 
ured by the number of terms that precede the unbroken 
run of zeros. 

Thus a quadratic (a, b, c, 0, 0, 0, • • •) where c ^ 0 is 
less simple than a linear polynomial (d, e, 0, 0, 0, • • •) 
where e ^ 0, and this in turn is less simple than 
(/, 0, 0, 0, • ■ • ) where f 0, and this is less simple than 
(0, 0, 0, 0, 0, • • •)? the polynomial zero. No polynomial 
simpler than zero can be found. 

If B(x) is any polynomial other than zero, the division 
process gives us R(x), of the form A(x) — Q(x)B(x), and 
this R(x) is simpler than B(x). 

Finally, if V\(x) is any given polynomial, we cannot 
find an unending sequence V\(x), V 2 &), Vz(x), • • • , with 

Analogy Between Integers and Polynomials 


every V n (x ) simpler than F n _i(x). For example, if V\{x) 
had 100 terms before the zeros, V 2 (x), being simpler than 
V\{x), could have at most 99. V%(x) could have at most 
98. (Note that V„(x ) is required to be simpler than V n -i(x). 
Its complexity must be less. “Less than or equal to” is not 
good enough.) So continuing, we see that V n (x ) has at 
most (101 — n) non-zero terms, until we reach Fioo(*) 
which has one non-zero term, and Vm(x) which is zero. 
At this point, the sequence must terminate. 

Requirements (1), (2), and (3) are thus met, and we 
can find the H.G.F. of two polynomials by a procedure 
closely akin to that for two integers. 

In the example below, we find the H.C.F. of 
x 4 — x 3 — 2x 2 + 2x — 4 and x 5 — 2x 3 — 3x 2 + x — 6. 
It so happens that the stages of this process run exactly 
parallel to our earlier calculation, step for step, so that 
comparison is particularly easy. 

X + 2 

x* — x 3 — 2x 2 + 2x — 4 

X s — 2x 3 — 3x 2 + x — 6 

x 4 — x 3 — 3X 2 + 4x — 4 

x 8 — 3x 3 — 2x — 4 


x 2 — 2x 

x 3 - 3x 2 + 3x - 2 

x 2 — 2x 

x 3 - 3x 2 + 2x 


x - 2 

The above example is very simple arithmetically, only 
integers being involved as coefficients. But it is not always 
possible to work without fractions appearing, as the 
next example shows. It was for this reason that we con- 
sidered in this section polynomials with rational coeffi- 
cients rather than polynomials with integral coefficients. 

i* + ! 

X 3 + 4x 2 + 4x + 3 

2x 3 + 5x 2 - lOx - 21 

x 3 + 4x 2 — 3x — 18 

2x 3 + 8x 2 + 8x + 6 

7x + 21 

- 3x 2 - 18x - 27 

- 3x 2 - 15x - 27 


92 A Concrete Approach to Abstract Algebra 

This calculation gives the H.C.F. as lx + 21. As this 
is 7(x + 3), we may remove the “unit” 7, and take x + 3 
as a correct answer. In fact, 

x 3 + 4x 2 + 4x 4* 3 = (* + 3)(x 2 + x + 1), 

2*3 + 5*2 _ 10 * _ 21 = (x + 3)(2x 2 - x - 7). 

We have now shown an analogy between integers and 
polynomials in regard to division, factors, primes, units, 
H.C.F. For we have explained all of these things in gen- 
eral terms that apply equally well to the two structures 
and, no doubt, to other structures also. 

There remains one last point to clear up. How many 
elements h can satisfy the conditions? 

(i) A is a common factor of a and b. 

(ii) h — au + bv for some u, v of the structure. 

(iii) Any common factor of a, b is a factor of h. 

Suppose that h and k both satisfy these. Then, in virtue 

of (i), A; is a common factor of a and b. By (iii), k must be 
a factor of h. .*. h — ek for some e. Similarly, by inter- 
changing the roles of h and k, k — fh for some /. Hence 
h = ek = efh. Therefore h{ 1 — ef) = 0. h is not zero, so, 
by axiom (72), 1 — ef = 0. Hence ef — 1. If m is any 
element of the structure, efm = m. That is, e and / are 
factors of every element of the structure. Accordingly, e 
and / are units. 

For the integers this means that for a = 12,£ = 18, 
the conditions (i) through (iii) are satisfied by 6 and — 6 
and by no other numbers. 

For the polynomials, if A(x) = ( x — 1)(* + 1) and 
B( x ) — ( x ~ 1)(* + 2), then the conditions for H(x) 
are satisfied by C(x — 1) for any constant C, and by no 
other polynomials. 

This degree of uncertainty causes us no trouble. We 
have already seen that multiplication by a unit is a 
trivial change. If we know that numbers a and b both 

Analogy Between Integers and Polynomials 


divide by 6, then we know that they both divide by — 6. 
It is quite easy to show that if h satisfies properties (i)- 
(iii), so does hf, where /is a unit. This degree of vagueness 
in the answer is therefore unavoidable. And we proved 
that this is all the vagueness there is. We did this when 
we showed that any k satisfying properties (i)-(iii) must 
be of the form hf. 

We can, if we like, make a convention to fix h. For 
numbers we can say that h must be positive; we choose 6, 
rather than —6, for the H.C.F. of ±12 and ±18. For 
polynomials, we can require the coefficient of the highest 
power of a; to be 1 . 

We shall exploit the analogy between integers and 
polynomials in the next chapter. 

Chapter 5 

An Application of the Analogy 

We have met two procedures for obtaining a new 
structure from an old one. 

First, given any field F, we have shown how to con- 
struct polynomials over that field. That is, given a field F, 
we show that it is always permissible to introduce a new 
symbol x and to assume that polynomials a + bx + 
cx 2 + • • • + kx n , where a, b, c, • • • , k are elements of F, 
satisfy the commutative, associative, and distributive 

Second, given a structure such as the integers, we obtain 
a field from it in the following way. We select a prime 
element, p , of the structure, and we break the structure 
up into classes. Two elements belong to the same class 
if they leave the same remainder on division by p. If 
element a belongs to class A and element b to class B , 
then A + B means the class containing a + b, and AB 
the class containing ab. By a structure “such as the 
integers,” we understand one in which all the field ax- 
ioms except (70) apply, and also the three conditions 
listed in chapter 4 that make the H.C.F. process possible. 
We then know that the classes A, B, • • • , form a field, for 
our proofs in chapters 3 and 4 used only the field axioms 
and the three conditions. 

An Application of the Analogy 


Both procedures we have a right to carry out. We do 
not need to have any doubts whether what we are doing 
is justified. We spent a good deal of time in chapter 2 
showing that a new symbol x could always be brought 
in : it expressed merely the properties of sequences 
(a, b, c, • • • , k, 0, 0, 0, • * •) formed from the elements of 
the field F. That is, it expressed the properties of some- 
thing already there: it did not really bring in anything 
new. In the same way, the second procedure dealt with 
classes of elements in the structure. It, too, involved no 
new assumption. 

It will be convenient to have names for these proce- 
dures, so that we can refer to them without a long ex- 
planation. The first procedure we will call adjoining an 
indeterminate x to the field F. In the second procedure, from 
a suitable structure S, we form the residue classes modulo p. 
Thus, the arithmetic of Even and Odd is the structure 
formed from the integers by considering the residue 
classes modulo 2. “Residue” is a word more or less 
synonymous with “remainder”; for some reason, it has 
become customary to use it in this connection. 

Having these two procedures, we naturally look for 
structures to apply them to: in this way, we hope for a 
good crop of new structures. 

The first procedure we have already applied to quite 
a number of fields: we have considered polynomials 
with real coefficients, polynomials with rational coeffi- 
cients, polynomials with coefficients O, I, II, III, IV 
from the arithmetic modulo 5, and polynomials with 
coefficients from the other modular arithmetics. It is also 
possible to adjoin an indeterminate x to a structure that 
is not a field, as when we consider polynomials with 
integers as coefficients: the integers do not form a field. 
This however we have not investigated yet. 

It does not look as though the first process is likely to 


A Concrete Approach to Abstract Algebra 

open new horizons for us. We have already used it quite 
often; it has not led to any particularly novel idea. 

How about the second process? So far, we have applied 
it to one structure only — the integers, from which we 
have obtained finite fields: the arithmetic of Even and 
Odd, the arithmetics modulo 3, modulo 5, modulo 7, 
and so on. To what else can we apply it? There is an 
obvious candidate. In chapter 4 we saw that polynomials 
resembled the integers very closely. The natural structure 
to consider is that of polynomials. This still leaves us 
some choice; we might consider polynomials with real 
coefficients, or rational coefficients, or coefficients from 
one of the modular arithmetics. 

Suppose we pick polynomials with real coefficients. 
When we obtained the arithmetic modulo 5 from the 
integers, our first step was to choose the prime number 5. 
Here, our first step must be to choose a prime poly- 
nomial — one that cannot be factored, x 2 + 1 is such a 
polynomial. For suppose it had a factor. This factor 
would have to be linear, say bx — c. Taking out the con- 
stant factor b, we can reduce bx — c to the form x — a. 
By the remainder theorem, x — a is a factor of a poly- 
nomial /(*) if and only if f(a) = 0. So x — a is a factor 
of x 2 + 1 only if a, 2 + 1 is zero. But, for any real number 
a, a 2 + 1 is larger than 1, hence not zero. So x 2 + 1 has 
no factors within the set of polynomials with real coeffi- 
cients. It is a prime polynomial. (In books on algebra 
the word “irreducible” is usually employed instead of 
“prime.” The meaning is the same.) 

Accordingly, if we begin with polynomials over the 
real numbers and form the residue classes modulo 
(x 2 -f- 1), we are bound to arrive at a field. 

When we constructed the arithmetic modulo 5, we 
only needed the symbols O, I, II, III, IV because divi- 
sion by 5 could only give the remainders 0, 1, 2, 3, 4. 

An Application of the Analogy 

9 7 

Division by a 2 + 1 will never give as a remainder any- 
thing more complicated than a linear polynomial, ax + b. 

These remainders serve to label the various classes; one 
class consists of all the polynomials that leave remainder 
4a: — 1, and so on. A remainder may of course be con- 
stant. Thus 4a: 2 + 6 leaves remainder 2. 

We introduce the following notation. ■ " 

Let 1 stand for the class of polynomials that leave remainder 1 . 

Let 2 stand for the class of polynomials that leave remainder 2. 

Let 3 stand for the class of polynomials that leave remainder 3. 

Let k stand for the class of polyno mials that leave remainder k. 

Let J stand for the class of polynomials that leave remainder x. 

This notation will prove sufficient for our needs. 

Since * belongs to the class J, and 3 belongs to the 
class 3,3J is the class containing 3a. 

Since 3 a: belongs to the class 3J, and 2 belongs to the 
class 2, 3J + 2 is the class containing 3x + 2. 

In the same way, quite generally, aj + b is the class 
containing ax + b. 

We could now, if we liked, work out sums and products 
directly from the definition. 

For example, to find (2J + 3)(4J + 5), we could rea- 
son that 2 J + 3 is the class containing 2a: + 3, and 
4J + 5 is the class containing 4a + 5. So(2 J + 3)(4J + 5) 
is the class containing (2a + 3) (4a + 5). (2a + 3) (4a + 5) 

= 8a 2 + 22a +15, which, on division by a 2 + 1, leaves 
22a + 7. The class containing 22a + 7 is 22J + 7. 


(2J + 3)(4J + 5) = 22J + 7. 

But it is not really necessary to go through all this. 

For we know that the residue classes constitute a field. 
Accordingly, they obey the ordinary laws of algebra. 

So we may multiply out i 

(2J + 3)(4J + 5) = 8J ! + 22J + 15. ; 

Two points arise here. First, are we entitled to as- - 1 


A Concrete Approach to Abstract Algebra 

sume that 2-4 = 8? Is it correct to multiply the bold- 
face numbers just as if they were ordinary numbers? 
Second what is J 2 ? 

The first question asks us to justify what we have 
already done. The second points to still unexplored 

The first point is soon dealt with; it does require us 
to go back to the definition, but as the result is a general 
rule, we are saved the bother of appealing to the defi- 
nition each time. 2 is the class containing 2, 4 is the class 
containing 4. By definition 2-4 is the class containing 
2-4, which is 8. The class containing 8 is called 8. 
So 2*4 = 8. 

The argument here used applies equally well to addi- 
tion; it does not in any way depend on the particular 
choice of the numbers 2 and 4. We conclude that the 
boldface numbers can be added and multiplied by the 
ordinary processes of arithmetic. We do not have to 
learn any new tables. In technical language, the ele- 
ments k are isomorphic to the numbers k. 

We now come to the second point: what is J 2 ? Here 
again, we go back to the definition of multiplication for 
residue classes. J is the class to which x belongs, so J • J 
is the class to which x-x belongs. That is, J 2 is the class 
containing x 2 . But when x 2 is divided by x 2 + 1 the 
remainder is —1. Hence J 2 = —1. 

We thus reach a striking conclusion. The elements 
a J + b can be added and multiplied by the ordinary 
rules of arithmetic and algebra together with the equa- 

J 2 = -I- 

We have, in effect, arrived at a theory of complex 
numbers. J does what we expect the square root of 
minus one to do. 

Let us look back at what we have done. We began 

An Application of the Analogy 


with the field of real numbers. We applied two proce- 
dures to it, that we knew were permissible and would 
lead to a field. We obtained a field, containing elements 
a, that behaved exactly like the original real numbers a, 
and also containing an element J for which J 2 = — 1. 

Since the elements a have exactly the same pattern 
as the numbers a , no harm will be done if we now forget 
the distinction between a and a. We can then say simply 
that to the real numbers a we adjoin a new element J 
for which J 2 = — 1 . The structure so obtained is a field : 
that is to say, our algebraic habits do not have to be 
changed when we are working with it. 

If you feel that there is something unfair in identifying 
a with a, it is quite possible to maintain the distinction 
between a and a and yet get information about the real 
numbers. We shall illustrate this by deriving a partic- 
ular identity, of some mathematical and historical inter- 

First of all, (a + bj)(a - bj) = a 2 - b 2 J 2 = a 2 + b 2 . 
Similarly, (c + dj) (c — dj) = c 2 + d 2 . Accordingly, 

(a 2 + b 2 )(c 2 + d 2 ) 

= (a + bj)(a - bj)(c + dj)(c - dj) 

= (a + bj) (c - dj)(a - bj)(c + dj) 

= {ac + bd + J(bc — ad)} (ac + bd — J(bc — ad)} 
= (ac + bd) 2 - J 2 (bc - ad) 2 
= (ac 4~ bd) 2 + (be — ad) 2 . 

But we showed earlier that the elements a, b, c, d 
combined with each other in exactly the same way as 
the numbers a , b, c, d. In view of this isomorphism, we 
deduce that, for any real numbers a, b, c, d, 

(a 2 + b 2 ) {c 2 + d 2 ) = {ac + bd) 2 + {be - ad) 2 . 

This result is of some interest in the theory of numbers. 
Any whole number can be expressed as the sum of four 
squares. Thus, 7 = 4 + 1 + 1 + 1 = 2 2 + l 2 + l 2 + l 2 . 


A Concrete Approach to Abstract Algebra 

Thus it is easily seen that 7 cannot be expressed as 
the sum of fewer than four squares; 1 and 4 are the only 
squares small enough to be used. It is thus a definite 
property of a whole number if it can be obtained by 
adding less than four squares together. For example, 13 
is 9 + 4, the sum of two squares. 17, being 16 -f- 1, is 
also the sum of two squares. The identity above tells us 
that, when two such numbers are multiplied together, 
their product also is the sum of two squares. 13 X 17 is 
221. If we put a = 2, b = 3, c = 4, d = 1 in the identity 
we get 

13-17 = ll 2 + 10 2 , 

and this expresses 221 as the sum of two squares. 

The identity also has some relation to trigonometry. 

This identity could, of course, be proved by means of 
elementary algebra without bringing J in at all. This 
must be so, for otherwise we should have got a result 
by using J that was not true for the algebra without J. 
But the whole point of our earlier work was to show that 
the presence of J did not make any logical difference : 
given the possibility of the algebra without J, the possi- 
bility of the algebra with J was a logical consequence. 

Thus the introduction of J cannot lead to any extra 
results. We do not want it to; extra results would be 
wrong results. It can, however, lead to more convenient 
ways of obtaining known results. It can simplify proofs 
and illuminate theories. If you are sufficiently familiar 
with complex numbers, the proof of the identity given 
above is a natural one. It follows a train of thought that 
could lead you to the identity if you did not already 
know it.* 

In trigonometry it is well known that many results 
can be obtained far more quickly and easily by using 

* If w = a + bJ, z = c — dJ, the identity, in modulus notation, 
states |w| • |z| = \wz\. 

An Application of the Analogy 


\/—l than by any other method. As the trigonometrical 
functions belong to analysis rather than to algebra (they 
require infinite processes for their definition), we shall 
not discuss them here. They are mentioned only as 
showing that the right to introduce V / — 1 is a valuable 

For the present, it will be sufficient to consider our 
demonstration of the algebraic identity above. Someone 
might say that this particular result could be proved 
just as easily by elementary algebra. We are not con- 
cerned with this criticism. \/ — 1 has proved itself very 
fruitful in many branches of mathematics; its claim to 
usefulness is not based solely on this particular identity. 
We are defending not the usefulness but the correctness 
of our method. Our proof may not be the neatest proof; 
but is it a proof at all? 

“No,” says our critic, “of course it isn’t. You have a 
symbol J and you write J 2 — — 1. Well, that just is 
not so. You can’t find a number J such that J 2 = —1.” 

The interesting thing about this argument is that 
everything, except the conclusion, is correct. We do have 
a symbol J and we do write J 2 = — 1. It is also true that 
no number has its square equal to — 1. 

Our critic’s mistake is in supposing that these true 
statements in any way detract from our proof of the 
identity. They do not. We have at no point said that J repre- 
sented a number. All we assumed, in the various steps we 
took, was that J obeyed the ordinary laws of algebra. Our“ 
critic’s assumption (probably based on something his 
teacher told him, authoritatively, and without any evi- 
dence to back it up) is that algebraic symbols always 
represent numbers. 

It may help to summarize the rather lengthy process 
we have been through, if we present our argument in the 
form of a discussion with the critic. 


A Concrete Approach to Abstract Algebra 

Ourselves. I understand that you feel quite happy with 
calculations that only use real numbers, such as 2, — 3, 
3/4, V2, tt? 

Critic. Yes. 

Ourselves. And you are then quite happy with poly- 
nomials such as 2x — 3 or x 2 + 2, and you are able 
to calculate with these? 

Critic. Yes. 

Ourselves. In particular, you admit that it is possible to 
divide such polynomials by x 2 + 1 and see what re- 
mainder results. 

Critic. Yes, of course. 

Ourselves. So that, for instance, x 3 + x + 2 and 3x 2 + 5, 
on division by x 2 + 1, both leave the remainder 2? 

Critic. That is correct. 

Ourselves. And there would be nothing vicious then in 
saying that all polynomials with this property could 
have the label 2 attached to them? 

Critic. No. I cannot object to that. 

Ourselves. And we could attach the label J to any poly- 
nomial that left remainder x, and the label — 1 to any 
polynomial that left remainder —1? 

Critic. Certainly. 

Ourselves. Would there be any objection to recording 
in the form 2-x = (2x) the fact that when any poly- 
nomial labeled 2 is multiplied by any polynomial 
labeled x the product is always to be labeled 2x? 

Critic. No. That is simply a notation, and does not 
commit me to any new admission. 

Ourselves. And, similarly, we could explain the nota- 
tion 2 + x? 

Critic. Yes. 

Ourselves. And we could then check that all the labels 
a J + b, together with the signs + and • , obeyed the 
axioms (7) through (72) for a field? 

An Application of the Analogy 


Critic. I am willing to accept your word that anyone 
who did not find it too boring could do so. 

Ourselves. And by means of these axioms we could 
justify all the steps taken with the symbols a, b, c, d, J 
in our demonstration of the identity? 

Critic (after some thought). Yes; you have only handled 
these symbols in accordance with the rules allowed by 
axioms (7) through (72). 

Ourselves. And finally we arrived at a result, 

(a 2 + b 2 )(c 2 + d 2 ) = (ac + bd ) 2 + (be — ad) 2 , 
in which J does not appear? 

Critic. Yes. 

Ourselves. But we also noticed that there was a very 
close resemblance between operations on the classes 
a, b, c, d and calculations with the corresponding 
numbers, a, b, c, d. For instance, can you find any 
example where a + b = p but a + b is not p ? Or 
one where a • b = q and a • b is not q? 

Critic (after trying a few examples, and making some 
algebraic calculations). No, there are no such cases. 

Ourselves. And also if a — b = r, then it must be that 
a — b — r? 

Critic. Yes, certainly. 

Ourselves. Our algebraic identity is built up by a chain 
of additions, subtractions, and multiplications? 

Critic. Yes. 

Ourselves. So the corresponding identity must hold for 
a, b, c , d ? 

Critic. Yes, it seems so. 

Ourselves. In that case, we have proved our identity 
with the help of the symbol J, which does not repre- 
sent a number. And you have admitted the logic of 
our procedure. 

Critic. I feel I am being cheated, but I cannot say just 


A Concrete Approach to Abstract Algebra 

The idea of V —1 is still new enough for us to feel 
that it is strange. But in past centuries, it felt equally 
strange when the number "s/2 had to be brought in to 
describe the ratio between the lengths of the diagonal 
and the side of a square. Before that time, only rational 
numbers, p/q, with p and q integers, had been considered. 

It was proved by the ancient Greeks that there cannot 
be a rational number * for which x 2 = 2. For if there 
were, this number could be expressed as p/q where p 
and q have no common factor (since any common factor 
could be canceled). In particular, we may assume that 
p and q are not both even: if they were both even, we 
could keep on canceling 2 until an odd numerator or 
denominator appeared. We suppose all possible canceling 
has been done, and that (p/q) 2 = 2; that is, p 2 = 2 q 2 . 
The square of an odd number is odd. But p 2 , being 2q 2 , 
is even. Hence p cannot be odd. So p must be even. 
Hence q must be odd, as all possible canceling has been 
done. But, as p is even, we can write p — 2k. Then 
j b 2 — 2 q 2 shows that Ak 2 — 2q 2 or q 2 = 2k 2 . But q is odd, 
so q 2 is odd. 2 k 2 is even. 

We have arrived at the conclusion that an odd number 
equals an even one. This is a contradiction. So there 
must have been something wrong with the assumption 
made at the outset, that x 2 — 2 could have a rational 
solution. Since this assumption leads to an absurdity, it 
must be false. 

So an ancient Greek, who thought of numbers as con- 
sisting only of rational numbers, 3§, f, 5|, and so on, 
could have said, “No number has square equal to 2,” 
with just as much conviction as our critic who said, 
“No number has square —1.” 

Our procedure is equally suitable for convincing him. 
We begin with the field of rationals, which he admits. 

An Application of the Analogy 


(An ancient Greek would not have recognized negative 
numbers. We suppose this difficulty to have been over- 
come first.) We adjoin to this an indeterminate *, and 
thus obtain polynomials with rational coefficients. We 
then select the prime polynomial x 2 — 2. This is prime 
over the rationals, for if it had a factor x — a with a 
rational, then a 2 — 2 would be zero, and there is no 
rational number for which this happens. So, when we 
classify polynomials with rational coefficients in accord- 
ance with their remainders on division by x 2 — 2 (in 
other words, when we form the residue classes modulo 
x 2 — 2), we obtain a field. 

Once again, we introduce a notation. 

1 is the label for any polynomial leaving remainder 1. 

2 is the label for any polynomial leaving remainder 2. 

K is the label for any polynomial leaving remainder x. 

Then K 2 or K • K, is the label for the class containing 

x 2 . But x 2 , on division by x 2 — 2, leaves remainder 2. 
So the class K 2 is the class 2. 

K 2 = 2 

As division by x 2 — 2 leaves a linear remainder a + bx, 
all our labels are of the type 

a + bK. 

We can prove results for rational numbers with the 
help of K, just as we proved results for real numbers 
with the help of J. The steps of the following argument 
run exactly parallel to the steps used earlier to establish 
an identity about the sums of squares. 

(a + bK)(a - bK) = a 2 - b 2 K 2 = a 2 - 2b 2 . 

(c + dK)(c - dK) = c 2 - 2d 2 . 



A Concrete Approach to Abstract Algebra 

(a 2 - 2b 2 ) (c 2 - 2d 2 ) 

= (a + bK) (a - bK)(c + dK)(c - dK) 
= (a + bK) (c - dK) (a - bK)(c + dK) 
= {ac - 2b d + K(bc - ad)} 

{ac — 2bd — K(bc — ad)} 

= (ac - 2bd) 2 - K 2 (bc - ad) 2 
= (ac - 2bd) 2 - 2 (be - ad) 2 . 

As before, there is an isomorphism relating a, b, c, d, 
to a, b, c, d , and we have, for rational numbers a , b, c, d, 
the relationship 

(a 2 - 2b 2 ) (c 2 - 2d 2 ) = ( ac - 2 bd) 2 -2 {be - ad) 2 . 

Once the logical correctness of introducing K with 
K 2 = 2 has been admitted, we may use the sign V2 
instead of K; and finally, in view of the isomorphism 
between the classes a and the numbers a , we may drop 
the boldface. We thus reach our usual notation a + bV 2. 


1. Show that, over the arithmetic modulo 3, the poly- 
nomial x 2 + 1 is prime (irreducible). Can a theory of complex 
numbers be constructed for the arithmetic modulo 3? Investi- 
gate and discuss. 

2. Show that x 2 + 3 is an irreducible (that is, nonfactor- 
able) polynomial over the arithmetic modulo 5. Is x 2 + 1 irre- 
ducible over the arithmetic modulo 5? Is a field obtained by 
applying the usual rules of arithmetic and algebra (i) to the 
elements a + bV — 1 , (ii) to the elements a + bV — 3? a, b 
stand for the elements 0, 1, 2, 3, 4 (or O, I, II, HI, IV, if 
you like) of the arithmetic modulo 5. 

3. In the arithmetic modulo 2 it is possible to write down 
all the factorable quadratics, by listing all the expressions 

All Application of the Analogy 


(x -f- a)(x + b) and multiplying them out. Do this. Is there 
any irreducible quadratic x 2 + px + q over this arithmetic? 
If so, how many such quadratics are there? 

4. Consider the residue classes, modulo x 2 + x + 1, of 
polynomials over the arithmetic modulo 2. Let 0 label the 
polynomials giving remainder 0, 1 label the polynomials giv- 
ing remainder 1, M label the polynomials giving remainder x. 
Fill in the spaces of the multiplication table below. 

0 1 M M + 1 




M + 1 

5. In question 4, let “modulo x 2 + x + 1” be replaced by 
“modulo x 2 + 1,” the question otherwise remaining unaltered. 
Answer this amended question : Do the residue classes, modulo 
x 2 -f- 1, of polynomials over the arithmetic modulo 2, form a 
field? If not, which axioms fail? 

6. Do the elements 0 , 1, M, M + 1 of question 4 form a 

7. Consider the residue classes, modulo x 2 , of polynomials 
over the real numbers. Let Q label the class of polynomials 
that leave remainder x, K the class that leave remainder k, 
for any constant k. Find the product (2 + 3Q) • (4 + 5Q). 
Do the elements a + bQ, form a field? 

8. Show that the H.C.F. procedure applied to a + bx and 
x 2 + 1 enables us to find a linear polynomial f(x) and a con- 
stant k such that 1 = (a + bx)f(x) + ( x 2 + \)k. Find the poly- 
nomial f(x) and deduce the reciprocal of a + bj where 

J 2 - -1. 

9. What field results from considering the residue classes 
modulo x of polynomials over the reals? If f(x) and g(x) are 
two polynomials belonging to the same residue class modulo x, 
what can be said about the graphs of y = fix) and y = g(x)? 

10. Determine the equation whose roots are the elements 
0 , 1, M, M + 1 of question 4. 


A Concrete Approach to Abstract Algebra 

Must We Distinguish Between 1 and 1 ? 

This is a point that troubles some students. In the 
ordinary course of development, we begin with numbers 
0, 1, 2, 3, • • • , as used in counting, the natural numbers. 
We then meet fractions, which we think of as lying be- 
tween the natural numbers; If, If, If and many others 
lie between 1 and 2. 

Then perhaps we meet negative numbers, —1, —If, 
— If, —If, —2, • • • , which we may think of as lying to 
the left of zero. 

With the Pythagorean Theorem we meet s/2, which 
lies between If and 1^; from the circle we meet tt, 
which lies between 3 and 3f. These are irrational num- 

Rational and irrational numbers together make up the 
real numbers. Then we go beyond these to consider 
complex numbers, x + jyV — 1. If y happens to be zero, 
the complex number reduces to a real number. We think 
of 4 + oV — 1 as being simply the real number 4. 

Thus real numbers are thought of as being part of the 
complex numbers; rational numbers as part of real 
numbers; integers as part of rationals; the natural num- 
bers as part of the integers. All this we think while in a 
state of innocence. 

Then we meet a mathematical philosopher who points 
out that, in order to justify the use of V : 2, we had to bring 
in a symbol K, a label for any polynomial that left 
remainder x on division by x 2 — 2. K thus stood for a 
collection of polynomials. In the same way 2 stood for 
a collection of polynomials — all those that leave re- 
mainder 2 on division by x 2 — 2. Then we had K. 2 = 2. 
Now, says the philosopher, the number 2 is a very differ- 
ent thing from 2, a collection of polynomials. Admittedly 

An Application of the Analogy 


there is an isomorphism, 2 + 3 = 5 corresponding to 
2 + 3 = 5. But 2 and 2 are different things, and you 
must not forget this. 

Now this matter is somewhat crucial, for an argument 
of this kind arises at each step in the extension of number. 
In passing from whole numbers to fractions, a philosoph- 
ical theory considers number pairs ( p , q) which eventually 
turn out to play the role of p/q. Now a number pair is 
not the same thing as a number, so (says the philosopher) 
we must distinguish between the natural number 4 and 
the fraction (4, 1) or 4/1. 

Negative numbers also can be introduced by number 
pairing. Thus +4 corresponds to any pair such as (7, 3) 
in which the first number is 4 more than the second, — 4 
to any pair such as (5, 9) in which the first number is 
4 less than the second. Thus the philosopher distinguishes 
between the natural number 4 and the positive inte- 
ger +4. 

In the same way a distinction is drawn between the 
integer +4 and the real number 4, and between the real 
number 4 and the number 4 + 0 V — 1 occurring in 
complex number theory. 

Thus, according to this philosopher, the real numbers 
do not form part of the complex, nor the rationals of the 
reals, nor the integers of the rationals, nor the natural 
numbers of the integers. 

How are we to decide what value to attach to such 
philosophical arguments? The correct procedure is, I 
think, indicated in a passage of a well-known children’s 

The first person he met was Rabbit. 

“Hallo, Rabbit,” he said, “is that you?” 

“Let’s pretend it isn’t,” said Rabbit, “and see 
what happens.” 



A Concrete Approach to Abstract Algebra 

Rabbit here seems to be taking a thoroughly scientific 
position. Let us apply his approach to the point at issue. 
Suppose you cling to the viewpoint of innocence, that 
the natural numbers, the integers, the rationals, the 
reals, the complex numbers are like a set of boxes, each 
contained in the following one. To what errors will this 
supposition lead you? So far as I can see, it will not lead 
you to any. 

Nor, on the other hand, do I see that the philosopher 
who insists that +4 is different from 4 is necessarily 
going to be led into any particular error. The whole 
question seems to be about ways of looking at things, 
rather than about any point of fact. We often have 
difficulty in saying whether or not two things are “the 
same.” Is the man of seventy the same person as the 
boy who used to attend the village school? In one sense, 
yes; in another sense, no. 

This difficulty is particularly acute when we are talk- 
ing about abstract ideas, when we are trying to decide 
whether the natural number 4 is the same as the fraction 
4/1 or different. 

Suppose you say, “The conductor stopped the band 
after it had played four bars.” Which “four” are you 
using there? Since you can count the four bars, it would 
be reasonable to answer, “The natural number four, the 
four we use when we count 1, 2, 3, 4.” But suppose the 
conductor had intervened a little later, when the band 
had played four and one-half bars. A\ is clearly a frac- 
tion, a rational number. This suggests that the 4 in “four 
bars” could be regarded as a rational number equally 
well. My own feeling is that it does not matter which 
you call it, that the distinction between natural number, 
positive integer, rational number, real number, cannot 
be maintained in such situations — in fact, that there is 
a lot to be said for the viewpoint of innocence. It would 

An Application of the Analogy 


be most unconventional to say that the band played for 
4 + OV^ — 1 bars, but I do not know that anyone could 
say you were actually wrong if you did this. So far as I 
can see, it passes Rabbit’s test: no disaster would overtake 
you if you did it. 

What do we mean by saying that two mathematical 
systems are “the same”? Consider some examples. If 
you had two calculating machines that were identical 
except that the numbers on one were written in black, 
on the other in blue, would you say they represented the 
same system? Everyone, I imagine, would say “Yes.” 
Next, suppose that the two machines gave the same 
results, but that their internal mechanisms were differ- 
ent — one perhaps using gear wheels, the other being 
electronic. Would you still say they embodied the same 
system? My own answer would be “Yes.” 

Now suppose we have two calculating machines. The 
first has keys marked in black, and it carries out addition 
and multiplication for the tables shown here. Call this 
the black system. 


M + I 

O I M M + I 

0 I M M + I 

1 0 M + I M 

M M + I O I 

M + I M I 0 


M + I 

0 1 M M + I 

O I M M + I 

0 M M + I I 

0 M + I I M 

We also have the blue system, a machine with keys 
marked in blue, that performs operations for the tables 
shown here, 


A Concrete Approach to Abstract Algebra 













You will notice that the top left-hand corners of the 
“black” tables are the same as the “blue” tables. Shall 
we say then that the blue system is a part of the black 
system? If we accept the earlier contention, that differ- 
ences of color and of internal mechanism are unim- 
portant, we are bound to answer “Yes.” The blue ma- 
chine is not, of course, part of the black machine, but 
the mathematical system represented by, or embodied 
in, the blue machine is part of the mathematical system 
represented by the black machine. 

Now the blue system is, of course, the arithmetic 
modulo 2, and the black system is the field obtained 
from that arithmetic (as in question 4, p. 107) by con- 
sidering residue classes for the irreducible polynomial 

AT 2 -f- AT + 1- 

Our philosopher arrived at these systems in that 
order. He first had the blue O and I standing for Even 
and Odd. Then he brought in the black 0 as a label for 
any polynomial over the arithmetic modulo 2 that hap- 
pened to be exactly divisible by x 2 + x + 1 (which you 
can write lx 2 + /* + /, with blue Z’s, if you like). The 
black /, the black M, and the black M + I were given 
similar meanings. 

Having arrived at the black system in this way, our 
philosopher cannot get it out of his head that the black O 
has a different meaning from the blue 0 . He maintains i 
that the blue system is not a part of the black system, 
because of this difference of meaning. 

Now we are grateful to him for pointing out that the 
black system can be derived, built up, from the blue 
system. That is a most helpful observation, and we shall 
often make use of it. But because this is “a” way of doing 

An Application of the Analogy 


things, it does not follow that it is “the” only way of 
doing things. In mathematics, it very often happens that 
a complicated structure is discovered before a simple 
one. The black system might have been discovered be- 
fore the blue one. Then, we should have discussed it in 
its own right. We should have checked to see that it 
satisfied all the axioms for a field. We could do all thi s 
without mentioning the blue system, without even being 
aware that the blue system was possible at all. Then 
someone might have noticed that the northwest corners 
of the black tables already constituted a closed system, 
a field, with elements O and I. The blue system would 
then have been discovered, and thought of, as a part of 
the black system. 

In the same way, we arrived at the complex numbers 
earlier in this chapter by considering the residue classes 
modulo x 2 -f- 1. This is one way of obtaining the com- 
plex numbers, but there are others. One can derive 
p + q\Z~ 1 by considering the number pair (p, q ) ; in an- 
other treatment, “s / — 1 is related to rotation through a 
right angle; in yet another, matrix theory is used. We 
do not regard these as different complex number systems, 
but rather as different ways of realizing one and the 
same pattern. It is a waste of time to argue whether y/ — 1 
“really is” a residue class of polynomials, or the number 
pair (0, 1), or a rotation through 90°, or a particular 

I do not believe that any mathematical result at all 
comes from the discussion in the last few pages. I have 
included this discussion for two reasons. First, we nat- 
urally feel that something questionable is being done 
when we “identify” 1, standing for a collection of poly- 
nomials, with the number 1. Second, students pick up 
wisps and echoes of the philosophical discussions, on 
whether the integer 4 and the rational number 4/1 are 


A Concrete Approach to Abstract Algebra 

the same or different, and are bothered by these. Of one 
thing I am certain: it is a mistake to lose any sleep over 
such matters. The question is psychological rather than 
philosophical or scientific. You should feel perfectly free 
to adopt whichever answer allows you to work most 
happily: indeed, if it suits you, you can adopt now one 
view, now the other, according to the problem you are 
working on. 

Chapter 6 

Extending Fields 

In CHAPTER5we considered a way of extending a given 
field. The method used the idea of residue classes, but, 
as we saw in a lengthy discussion, it was often incon- 
venient to keep the final result in terms of residue classes. 
The residue classes gave us a way of showing that a 
certain type of calculating machine could be built, and 
indeed of building it. However, once having obtained the 
new machine, we shall often want simply to work it, 
and to forget what is inside it. So let us look at the fields 
we obtain by the procedure of chapter 5, and see just 
what this procedure does for us. 

At the beginning of chapter 5, by considering residue 
classes modulo (x 2 + 1), we obtained the symbols a + bJ 
where a and b are real numbers and J satisfies the equa- 
tion J 2 = —1. a bJ is commonly referred to as a 
“complex number.” In order to work correctly with 
complex numbers, all you need to know is (i) that 
complex numbers obey the laws of algebra, (ii) that 
J 2 = — 1 . Statement (i) here could be put, that complex 
numbers form a field. As the real numbers form a field, 
in passing from real numbers to complex numbers, we 
are not conscious of any change, so far as statement (i) 
goes. The main novelty lies in statement (ii), that 
J 2 — —1. So long as we are working with the real 


A Concrete Approach to Abstract Algebra 

numbers, the equation x 2 + 1 =0 has no solution. The 
effect of passing to the complex numbers is to bring in a 
new symbol J, such that J 2 + 1 =0. We thus, so to 
speak, create a root for the equation x 2 + 1 = 0. You 
will notice that the procedure of chapter 5 used residue 
classes modulo (x 2 + 1). 

In the same way, toward the end of chapter 5, to 
obtain a field in which x 2 — 2 = 0 had a root, we con- 
sidered residue classes modulo ( x 2 — 2). 

Quite generally, if f(x) is an irreducible polynomial 
over a field F, we can obtain a field in which the equa- 
tion f(x) = 0 has a root by considering residue classes 
modulo /(*). 

(You should be able to prove this result, by observing 
the proofs on pages 98 and 105, for J 2 + 1 =0 and 
K 2 — 2 = 0, and noting that the method used in these 
two particular proofs can be used in general. A proof 
will be given shortly, but it does no more than carry 
out the hint here given : if you can find the proof unaided, 
you will gain in insight and confidence.) 

In speaking above of statements (i) and (ii) as em- 
bodying all you need to know to calculate with complex 
numbers, I should perhaps have added something you 
need to know is not so. A student might write 2 J — 3 = 0, 
and say “Why not? How do you know this is not so?” 
In this particular case, it is easy enough to show that 
this equation is wrong. If 2J — 3 = 0, J = 1§, so 
J 2 — — 1 would mean (1|) 2 = — 1, which is not so. 

However, with an eye on the general moral we are 
hoping to draw, we might state (ii) more fully ; J 2 = — 1 
and J does not satisfy any simpler equation (that is, 
any equation of lower degree). We are now ready to 
state our general theorem. 

theorem. If F is any field, and f{x) is an irreducible 
polynomial over F, we can always construct a new field , con- 

Extending Fields 


taining the elements of F and also a new symbol Q for which 
f(Q) = 0 but Q does not satisfy any equation of lower degree. 

Residue classes are not mentioned at all in the state- 
ment of this theorem, but of course we go back to res- 
idue classes for our proof that the new field can be con- 
structed. However, when we apply this theorem, we need 
not think of residue classes at all, unless we want to. 

Proof. It will be convenient to write the proof out on 
the assumption that f(x) is a cubic. It should be clear 
that the ideas of the proof work equally well for a poly- 
nomial of any degree. 

If k is any constant, the equations /(Q) = 0 and 
kf(Q) = 0 are equivalent; /(*) being irreducible and 
kf(x ) being irreducible are also equivalent. Thus we do 
not lose any generality if we suppose f(x) to have the 
coefficient 1 for the highest power of x. 

Suppose then that f{x ) is x 3 — ax 2 — bx — c. We con- 
sider residue classes of polynomials over F, modulo f(x) . 
Since f(x) is irreducible, these residue classes will form a 

Let a stand for the class containing a. 

Let b stand for the class containing b. 

Let c stand for the class containing c. 

Let Q, stand for the class containing x. 

You will remember that addition and multiplication 
of classes were defined by means of representatives. For 
instance, in the arithmetic modulo 5, we found III • IV 
by taking 3, an element in class III, and multiplying it 
by 4, an element in class IV. 3-4 = 12, and 12 is in 
class II, so III -IV = II. We showed (in chapter 3) that 
it did not matter which representatives we took. We 
might, for example, have chosen 8 to represent class III 
and 19 to represent IV. 8*19 = 152, and 152 is in 
class II, so we still reach the result III -IV = II. 

We are now working with polynomials, but (as chap- 


A Concrete Approach to Abstract Algebra 

ter 4 stressed, and chapter 5 applied) polynomials can 
be handled much like integers. We thus find 

Q? is the class containing x z . 
aQ 2 is the class containing ax 2 . 
b Q is the class containing bx. 
c is the class containing c. 

Combining these, we see that Q? — aQ 2 — bQ — c is 
the class containing x s — ax 2 — bx — c. But x 3 — ax 2 — 
bx — c is /(x); on division by /(x) it leaves remain- 
der zero. Hence it belongs to the class 0. Therefore 
Q? — aQf — bQ — c and 0 are labels for the same 
class, that is 

Q 3 - aQ 2 - bQ - c = 0. 

We now have to finish off the argument, as we did in 
chapter 5, to show that we could drop the boldface 
and write 

Q 3 — aQ 2 — bQ — c — 0, 

that is, 

AQ) = o. 

This is the main result we wanted to show. But we 
also have the negative result to prove, that Q does not 
satisfy any equation of lower degree. This comes very 
simply from considering residue classes. 

Suppose someone claimed that Q satisfied an equation 
pQ 2 + rQ + s = 0. I suppose here that p, r, s are not all 
zero. If someone says 0-Q 2 + 0- Q + 0 = 0, we can 
only agree with him. That is perfectly correct, but it is 
trivial; if you like, you can reword the theorem to 
exclude this case specifically. 

Suppose, then, it is claimed that pQ 2 + rQ + s = 0. 
In terms of residue classes, this means that pQ 2 + rQ + 
s = 0. pQ 2 + rQ + s is the class containing /w 2 + rx + s. 
The equation asserts that px 2 + rx + s leaves remainder 

Extending Fields 


0 on division by x z — ax 2 — bx — c. But this is so only 
if p, r, s are all zero, and we have already excluded this 
trivial case. 

We have now proved all the assertions contained in 
the theorem. Q.E.D. 

It will be convenient to make a remark about the new 
field while we still have in mind the symbols as defined 
in the proof of the theorem. We want to show that the 
facts given in the theorem completely replace the res- 
idue-class approach so far as making calculations is 

With f(x) cubic, it is clear, on the residue-class ap- 
proach, that we never need anything higher than a 
quadratic in Q, to write down an element of the new 
field. For the remainder, on dividing any polynomial by 
a cubic /(*), will be at most a quadratic in x, and its 
label will be at most quadratic in Q, How is this result 
arrived at with the new approach? 

The theorem does give us a way of getting rid of all 
powers of Q above the second. For it states /(Q) = 0, 
and this is equivalent to 

Q 3 = aQ 2 + bQ + c. (I) 

(It was in anticipation of this equation that f(x ) was 
defined as x 3 — ax 2 — bx — c, rather than with positive 

Accordingly, we have a quadratic expression that can 
always be substituted for Q 3 . If we multiply both sides of 
equation (I) by Q, we obtain 

Q 4 = aQ 3 + bQ 2 + cQ. (II) 

Substitute in (II) the value of Q 3 given by (I). We find 

Q 4 = a(aQ 2 + bQ + c) + bQ 2 + cQ 

= ( a 2 T" b)Q 2 T (ab -f - c)Q -f- ac. (Ill) 

It is evident that by continuing in this way, any 


A Concrete Approach to Abstract Algebra 

power whatever of Q can be expressed as a quadratic 
in Q. Hence any polynomial in Q can be reduced to a 

Thus, if we take two quantities of the form rQ 2 + sQ + t 
and add or multiply them, the result is always expressible 
in this form. 

We can show, too, that every such quantity other 
than zero has a reciprocal of the same type. Let g(Q) = 
rQ 2 + sQ + t. By the H.C.F. procedure, applied to the 
polynomials g(Q) and /(Q), we can find polynomials 
u(Q) and v(Q) such that 

g(QMQ) +/(QMQ) = 1. 

But /(Q) = 0. Hence 

g(QMQ) = l. 

u(Q) is a polynomial in Q. Actually, the H.C.F. process 
will yield a quadratic u(Q) and a linear v(Q). But we 
can save ourselves the trouble of proving this by means 
of the remark above, that any polynomial in Q can be 
reduced to a quadratic. 

Thus 1 /g(Q) can be expressed as a quadratic in Q. 

Division by g(Q) can be achieved by multiplying by 

1 /g(Q). 

These considerations are intended to illustrate the 
fact that the properties of the new field can be obtained 
from the information contained in the theorem alone — 
that is, by purely algebraic calculations, without any 
mention of residue classes or the meaning of Q. We 
appeal to residue classes to satisfy our scientific con- 
science, to show that what we are doing is justified. But 
once the foundations have been laid, we can proceed 
purely by calculation — as, of course, mathematicians 
did for centuries with V — 1 and V / 2 before the residue- 
class explanation was thought of. 

Extending Fields 


The field formed by adjoining the element Q to the 
field F is often denoted by F(Q). This, of course, has 
nothing to do with the symbol for function, such as /(*). 

For example, if R stands for the field of real numbers, 
R(y / — 1) will stand for the field obtained from R by 
introducing the new element y/ — 1; that is, R(V- 1) 
stands for the complex numbers. 

Question: Verify by calculation that the elements 
r V / 4 + s's / 2 + t (r, s, t rational numbers) can be added, 
subtracted, multiplied, and divided, the results always 
being expressible in the same form. (This is the particular 
case where /(x) = * 3 — 2, so Q = V2, Q 2 = V4.) Find, 
in this form, the reciprocal of 

's/a + 2^2 + 3. 

Two Finite Fields 

As an illustration of these ideas, we shall consider 
two fields obtained as extensions of the arithmetic mod- 
ulo 2. These fields contain only a finite number of ele- 
ments so that they are very compact and easily surveyed. 

The procedure, as explained earlier, is to select an 
irreducible polynomial /(*), and then introduce a new 
symbol, say Q, such that/(Q) = 0. 

We first choose an irreducible quadratic for /(*); 
later we study what happens when an irreducible cubic 
is chosen. The irreducible quadratic has already been 
touched on in various examples (pp. 106-107, questions 
3, 4, 6, 10). The discussion of the irreducible cubic is 
equally suitable for an example. After reading the ac- 
count of the irreducible quadratic here, you may well 
wish to work out the other question for yourself, and see 
what properties you can observe or prove in the field 
you obtain. 


A Concrete Approach to Abstract Algebra 

A Field with Four Elements 

We propose to join a new element, M, to the arithmetic 
modulo 2. M is to be the root of an irreducible quadratic. 
Our first step, then, must be to find an irreducible quad- 
ratic. We can easily list all the quadratics that are 
reducible. Any that do not appear on this list must 
necessarily be irreducible. 

The only possible linear factors are x and x + 1, 
since 0, 1 are the only elements of the arithmetic. The 
factors of the quadratic may be both x ; or both x + 1 ; 
or one of each. 

Thus, the only quadratics that factor are 

xx = x 2 

x - (* -f- 1) = x 2 + x 
(*+ l)-(*+ 1) = x 2 +l 

Every quadratic is of the form x 2 + px + q, where p 
may be 0 or 1, and q also may be 0 or 1. 

Thus, every possible quadratic appears in the scheme 

p = 0 p = 1 

q = o 

q = 1 

Three of these quadratics have already been listed as 
reducible. One only remains, x 2 + x + 1, and this is 
irreducible. Accordingly, we introduce M to provide a 
root for this polynomial. That is, we assume 

M 2 + M + 1 = 0. 

Since we are working in the arithmetic modulo 2, 
this equation is equivalent to 

M 2 = M + 1. 

Extending Fields 


(Justification. Add M 2 to both sides of the earlier equa- 
tion, and use 1+1=0 for the coefficient of M 2 on 
the left-hand side. M + 1 = M 2 is the result.) 

By means of this equation, we can always replace 
M 2 by M + 1, and thus we never need to have the 
square of M, or any higher power, left in an expression : 
we can always simplify until a polynomial is reduced to 
linear form. 

Thus, all our elements will be expressible in the form 
aM + b, where a and b take the values 0, 1. So we have 
the scheme: 

6 = 0 b = 1 

a = 1 M M + 1 

a = 0 0 1 

This shows that the new field contains only four ele- 
ments 0, 1, M, M + 1. 

The addition and multiplication tables for these four 
elements are : 




M + 1 





M + 1 




M + 1 




M + 1 



M + 1 

M + 1 







M + 1 










M + 1 




M + 1 


M + 1 


M + 1 



The addition table is easily obtained: we have only 
to remember, since we are working modulo 2, that 
1 + 1 and M + M are both zero. 


A Concrete Approach to Abstract Algebra 

In the multiplication table, we need to use the equa- 
tion M 2 = M + 1 . Thus 

M-M = M 2 = M + 1 
M-(M+ 1) = M 2 + M 

= (M + 1) + M = 1 

(Af + 1) • (M + 1) = M 2 + 2M + 1 = M 2 + 1 
= (M + 1) + 1 = M. 

The values thus found have been entered in the multi- 
plication table above. 

On seeing this multiplication table for the first time, 
students often comment on the pattern it displays — the 
stripes running from northeast to southwest. It is inter- 
esting to observe such a pattern. However, it is a mistake 
merely to notice a pattern, and pass on. 

Very often a mathematical discovery starts with an 
observation of some pattern. Then one asks, “What is 
this pattern? Why does it occur here? How can it be 
made to yield a mathematical theorem or principle?” 
After thought and analysis, one can often reach a result 
much more definite and useful than the original recog- 
nition of a superficial pattern. The next few paragraphs 
follow the lines of an actual class discussion. All the 
answers are those produced by the students. 

First of all, then, what is the pattern we have seen? 
If we leave out the border of zeros, the multiplication 
table has the following pattern. 

a b c 

b c a 


Where have we seen this kind of pattern before? In the 
arithmetic modulo n. In which table did it occur? In 
the addition table: the addition tables always showed 
this northeast-to-southwest striped effect. All the addition 
tables had this general type of pattern : did any particular 

Extending Fields 


table actually reproduce the identical pattern above? 
Yes: the arithmetic modulo 3. (See your answers to 
questions on page 15-16.) 

Our result is now rather more definite: the multiplica- 
tion table in our field reproduces the pattern of the 
addition table modulo 3. This is in itself quite striking. 
We have the following correspondence. 





M + 1 

modulo 3 




Multiplication in the first column corresponds to addi- 
tion in the second column. What does that remind us of? 
Logarithms. We could in fact use the table above exactly 
like a table of logarithms, for example, to multiply M 
by M + 1. Opposite M we find 1, opposite M + 1 
we find 2. Add 1 to 2. The sum (modulo 3) is 0. Now we 
decode (“take antilogarithms”). 0 stands opposite to 1. 
So the answer is 1 , which is correct. 

What, then, is a logarithm? The usual definition is 
that logb a — x if a = b x . A table of logarithms is just a 
table of powers, read the other way round. 

So our little table above suggests that there is some b 
such that 

1 = b\ 

M = b\ 

M+\ = b\ 

The middle entry here tells us what the base b of our 
log table is: b = M. In fact, then, 

1 = M\ 

M = M\ 

M + 1 = M\ 


A Concrete Approach to Abstract Algebra 

In our field, then, every element except zero is a power 
of M. 

Why should modulo 3 come into it? Let us continue 
to calculate the powers of M : 

M 3 = M-M 2 = M(M + 1) = M 2 + M = 1, 

M 4 = M'M 3 = M, 

M 5 = M-M 4 = M 2 , 

M 6 = M-M 5 = M 3 = 1. 

It is pretty clear now that, for any whole number n, 

M 3n = 1, 

M 3n+1 = M, 

M 3n + 2 = M 2 . 

Accordingly, if you want to find any power of M, say 
M k , where k is a whole number, all you need to know 
is whether k leaves remainder 0, 1, or 2 on division by 3. 
That is why modulo 3 comes into the picture. 

A modular arithmetic always comes in when we are 
dealing with roots of unity. For example ( — 1)* is 1 if 
k is even, — 1 if A; is odd. 

It is clear that sooner or later the powers of an element 
of the field must recur. Otherwise we could go on for- 
ever, forming M, M 2 , M s , M 4 , - • • , and getting new ele- 
ments all the time. But the field contains only four ele- 
ments; the supply of new elements must give out pretty 
quickly. This type of argument clearly applies to any 
field with a finite number of elements. 

Our inquiry naturally raises a number of general 
questions. Gan an irreducible polynomial f(x) of any 
desired degree k be found for the arithmetic modulo n 
(n prime)? 

If so, we can construct a field by adjoining an element 
M for which f(M) = 0 is assumed. Will all the elements 
of this field, apart from zero, be powers of M? Or is this 

Extending Fields 


property confined to the case we have just considered? 
Or does it hold sometimes but not always? 

The answers to these questions are known and can be 
found in standard texts, such as Van der Waerden’s 
Modern Algebra. You may derive some entertainment by 
collecting evidence bearing on these questions and try- 
ing to discover, or to prove, the answers for yourself. 
Some relevant evidence is provided later in this chapter, 
when we study a field with eight elements. 

It may be worth noting a curious symmetry between 
the elements M and M + 1 . 

M is a root of the equation x 2 -f- x -}- 1 =0. So is 

M + 1 . 

If you add 1 to AT you get M + 1 
If you add 1 to M + 1 you get M. 

If you square M you get M + 1 . 

If you square M + 1 you get M. 

Every element of the field, except 0 is a power of M. 
Every element of the field, except 0, is a power of 

This symmetry is significant and plays an important role 
in more advanced theory. 

A Field with Eight Elements 

We now begin to construct another field. We shall 
work much as we did on the field with four elements, 
except that we shall begin by seeking an irreducible 
cubic rather than an irreducible quadratic. 

There are eight cubics, since in * 3 + px 1 + qx + r each 
of p , q, r can be 0 or 1 . As before, we begin by listing all 
those that can be factored. A cubic can factor into a 
linear factor and an irreducible quadratic, or it may 


A Concrete Approach to Abstract Algebra 

break up completely into linear factors. The following 
list covers all the possibilities: 

x(x 2 + * + 1) = X 3 + X 2 + X, 

(x + 1)(* 2 + * + 1) = X 3 + 1, 

X’X’X = X 3 , 

x-x(x + 1) = X 3 + x 2 , 
x • (* + 1 ) 2 = X 3 + X, 

(x + l) 3 = X Z + X 2 + * + 1. 

Two cubics remain. They are x 3 -f x + 1 and x 3 + x 2 + 1 . 
These are the irreducible cubics over the arithmetic mod- 
ulo 2. We could choose either of them. Suppose we 
choose x 3 + x + 1. 

Our next step is to introduce Q with the assumption 
Q 3 + Q + 1 = 0, which is the same thing as saying 

Q 3 = Q + 1 . 

This equation shows that we need never keep Q 3 or 
any higher power in any expression. All our elements can 
be put in the form aQ 2 + bQ + c. Earlier we arranged 



Extending Fields 


the four elements of the other field systematically in a 
square (on p. 123). To arrange our present eight ele- 
ments systematically, we need a cube. 

The floor of the cube corresponds to a = 0, the ceiling 
to a = 1 ; the front to b = 0, the back to b = 1 ; the 
left to c = 0, the right to c = 1 . 

In our earlier work, the powers of M gave us all the 
non-zero elements of the field. Do the powers of Q do 
the same now? To answer this, we calculate the powers 
of Q. 

Q° = 1 , 

Q 1 = Q, 

Q 2 = Q\ 

Q 3 = Q + 1, our basic assumption, 

Q4 = Q • Q 3 = Q* + Q 5 

Q5 = Q.Q4 = Q3 + Q2 = Q2 +Q + 1j 
Q6 = Q ■ Q® = Q3 Q2 Q _ Q2 + 1} 

Q 7 = Q-Q 6 ~ Q 3 + Q = 1. 

(The last step in finding Q 5 , Q 6 , and Q 7 consisted in sub- 
stituting Q -f 1 for Q 3 , and simplifying, with the arith- 
metic modulo 2.) 

Q° through Q 6 are all different; they give us seven 
elements of the field: zero is the eighth. Once again all 
the non-zero elements are powers of a single element. 


1. What arithmetic modulo n plays the role for the field 
with eight elements that the addition modulo 3 played for the 
field with four elements? 

2. Every non-zero element can be represented as a power 
of Q. Is Q the only element with this property? If not, what 
other elements have it? 


A Concrete Approach to Abstract Algebra 

3. The equation x 3 + x + 1 = 0 has the root Q. Has it 
any other roots in the field with eight elements? 

4. The cubic x 3 + x 2 + 1 is irreducible over the arithmetic 
modulo 2. Is it irreducible over the field with eight elements? Do 
any elements of the field satisfy the equation x s + * 2 + 1 =0? 

5. If we had chosen x 3 + x 2 + 1 instead of x 3 + x + 1 for 
our irreducible f(x), would it have made any essential differ- 
ence to the work? Should we have arrived at the same field? 

6. Which powers of Q satisfy the equation x 3 + x + 1 = 0. 
Do you notice anything about these powers? 


Fields with a finite number of elements are known as 
Galois Fields, in honor of the young French mathemati- 
cian who first investigated them, rather more than a 
century ago. The letters GF are used as an abbreviation 
for Galois Field. 

GF( 2) is used to denote the arithmetic modulo 2, since 
this contains two elements 0, 1. The field with the 4 ele- 
ments 0, 1, Af, M + 1 is denoted by GF{ 2 2 ). The field 
with 8 elements is denoted by GF( 2 3 ). 

Quite generally GF(p n ) denotes a Galois Field with p n 
elements, where p is a prime number. (This is a slightly 
confusing abbreviation. In it, p n has its usual algebraic 
meaning, but GF must not be interpreted algebraically 
as a product. GF is simply an everyday abbreviation like 

Chapter i 

Linear Dependence and Vector 

In chapter 6, we noticed in passing that the elements 
of GF( 2 3 ) could be arranged in the shape of a cube, while 
those of GF( 2 2 ) could be arranged in a square. GF(2) 
of course has only the two elements 0, 1 and may be vis- 
ualized with the help of a line. 

0 1 

Here is a hint that fields may have some kind of 
geometrical aspect. 

In this chapter we develop some ideas that will prove 
very useful for the study of fields. The language of this 
chapter is geometrical. “Linear” is connected with the 
idea of “line”; “vector,” is derived from the Latin 
verb “to carry,” and is associated with the idea of mov- 
ing from one point to another. However, as was stressed 
in chapter 1 we are concerned with structures, not with 

We shall not confine ourselves to geometrical applica- 
tions, but we shall consider anything that has the same 
structure as certain geometrical entities. It is, in fact, 
very helpful to be able to pass quickly from one partic- 
ular representation to another, now using a geometrical, 


A Concrete Approach to Abstract Algebra 

now a numerical, now a physical realization of a struc- 
ture, so that we see the analogies between all of these 
but are never tied to any one of them. 

Let us start with a very physical example. You wish 
to buy some nails and screws in a hardware store. The 
storekeeper has observed that his customers usually buy 
such things for fitting hinges (using six screws) or for 
fixing cupboards to walls (using four screws and four 
nails). He has all his screws and nails done up in little 
packets, “The Home Handyman” with 6 screws, and 
“The Complete Carpenter” with 4 screws and 4 nails. 

If you want to buy screws and nails for some purpose 
other than those envisaged by the storekeeper, and you 
do not want to upset the store’s organization by asking 
to have the packets opened, there are certain limitations 
on what you can obtain. You can buy 26 screws and 
8 nails by taking 3 Handyman packets and 2 Carpenter 
packets, but you certainly cannot get 3 screws and 
15 nails. 

Any collection of screws and nails that you are able 
to buy (without splitting packets) we shall call a possible 
purchase or, more briefly, a purchase. We shall consider 
purchases in their algebraic aspect, and we shall also 
consider how to represent them geometrically. 

What algebraic operations are possible with purchases? 
The most obvious one is addition — the putting together 
of two purchases. Thus 5 Handyman and 2 Carpenter 
packets is a purchase; 3 Handyman and 7 Carpenter 
packets is another purchase. The sum of these two pur- 
chases is 8 Handyman and 9 Carpenter packets. 

There is no obvious way of multiplying a purchase by 
a purchase. A child would know what you meant by 
adding a Handyman packet to a Carpenter packet, but 
he would be puzzled to multiply them. 

There is, however, one kind of multiplication that is 

Linear Dependence and Vector Spaces 


possible. We know what is meant by one purchase being 
three times as large as another. If a purchase consists 
of 2 Handymans and 5 Carpenters, three times this 
purchase is 6 Handymans and 15 Carpenters. 

Thus we can add a purchase to a purchase, but we 
cannot multiply a purchase by a purchase. We can how- 
ever multiply a purchase by a number. This suggests a 
final question: can we add a number to a purchase? 
Imagine the following conversation. 

Clerk. Is there anything else? 

Customer. Yes. Three. 

Clerk. Three more Carpenters, sir? 

Customer. No. Just three. 

Clerk. Do you want a metal three to put on your front 

Customer. No. Just the number three. 

The clerk hopes it is a harmless lunatic he is dealing 

It is thus quite natural, in the situation described, 
to say 

Purchase -j- Purchase is defined. 

Number X Purchase is defined. 

Purchase X Purchase is not defined. 

Number + Purchase is not defined. 

How arbitrary this would look in abstract formula- 
tion! “We have two sets of objects. The set P contains 
elements pi, pi, pz, • • • ; the set M contains elements 

^lj ^3j • 

axiom 1. To any two elements pi, pi, of P there corresponds 
an element of P, denoted by pi + pi. 

axiom 2. To any elements pi of P and ni of M there cor- 
responds an element nipi of P. 

“We do not, however, define products of the type 
pipi, nor sums of the type m + pi.” Why not, you won- 


A Concrete Approach to Abstract Algebra 

der. The reasonableness of the above passage appears 
as soon as you recognize P as the set of all possible pur- 
chase, and M as the set of natural numbers. 

We now introduce a notation that is convenient for 
our discussion of purchase, and that will have other 
uses later. 

Let (x,y) stand for “x screws andy nails.” 

Let H stand for a Handyman packet and C for a 
Carpenter packet. We can then write 

H=( 6,0) (1) 

C = (4, 4) (2) 

For the purchase that we considered earlier, consisting 
of 3 Handymans and 2 Carpenters, we can write 

3H + 2C = (26, 8) (3) 

You can, of course, check this last equation by going 
back to its meaning. However, since we are interested 
in the formal aspects of this matter, we naturally try to 
obtain rules for calculation with expressions of this kind. 
We are guided, in making these rules, by what we know 
about the meaning of the symbols. 

Suppose we are given 

F - (a, b) (4) 

G = (c, i) (5) 

How shall we find F + G? We go back to the meanings 
of F and G. 

F stands for a screws and b nails. 

G stands for c screws and d nails. 

Putting these together, 

F + G stands for {a + c) screws and ( b + d) nails. 


F G — (a c, b d) 

( 6 ) 

Linear Dependence and Vector Spaces 


If we now look at equations (4), (5), (6), we can see 
that these equations give a formal definition of F + G. 
In the same way, if n is a number, we can show that 

ri'F — ( na , nb ) (7) 

This is easily justified in terms of screws and nails. But 
equations (4) and (7) themselves show by what rule nF 
is obtained from F; we can use this rule without making 
any reference to the meaning of F. 

Graphical Interpretation 

It is easy to illustrate what we have been considering 
with ordinary graph paper. Some possible purchases are 
shown in the following diagram, the number of screws 

12 , 





being measured horizontally and the number of nails 

You will notice that, starting from any point of this 
set that you like, adding H always takes you 6 units to 
the East. Adding C always takes you a certain distance 
(actually 4V2 units) to the North-East, 


A Concrete Approach to Abstract Algebra 

This shows the relation of vectors to the word “I 
carry,” and we could in fact explain vectors (as was 
done in the early days of vector theory) as representing 
displacements, (x, y) would then stand for “x inches to 
the East and y inches to the North.” 

Clearly, if operation F sends us a inches East and 
b inches North, while G sends us c inches East and d inches 
North, the combined effect of F and G (one operation 
followed by the other) is to send us ( a + c) inches East 
and ( b + d) inches North. Thus the addition procedure 
defined by equations (4), (5), (6) above still holds good. 
The operation F, repeated n times, will take us na inches 
East and nb inches North; so this ties in well with 
equation (7). 

With these operations we are no longer restricted to 
the natural numbers. We can have fractions and negative 
numbers, with the usual understanding that — 2| inches 
East means 2\ inches West. 

Suppose PQR represents a piece of wire. We displace 
this wire say an inch to the North-East. Every point of 


Linear Dependence and Vector Spaces 


the wire moves an inch to the North-East. P'Q'R' rep- 
resents the final position of the wire. The arrows show 
how each point moves. Any one of these arrows shows 
how the whole wire moves — provided we are told that 
every point undergoes the same displacement. Thus a 
displacement can be represented by an arrow. Usually 
we draw all our arrows starting from the same point, 0, 
as shown below. But of course this is only a convention. 

, // , . , 

I To North 

“A displacement of one inch to the North” means that 
every particle of a body moves one inch to the North. 
Suppose we have two displacements : 

F 1 inch to the East, 

G 1 inch to the North-East. 

C B 


A Concrete Approach to Abstract Algebra 

What is the combined effect of these? Consider F 
followed by G. F would send 0 to A. G would send A 
to B. The combined effect is to send 0 to B. 

If we considered G followed by F, then G would send 
0 to C. F would send C to B. The combined effect is 
again to send 0 to B. 

Thus the combined effect of the displacements repre- 
sented by the arrows OA and OC is the displacement 
represented by OB. It does not matter in which order 
the displacements OA and OC are combined. 

More generally, with 

F a inches to the East, b inches to the North, 

G c inches to the East, d. inches to the North, 

Linear Dependence and Vector Spaces 


the arrow OA represents the displacement F, the arrow 
OC represents the displacement G, the arrow OB repre- 
sents the combined effect. 

We denote the combined effect by F + G. The par- 
allelogram OABC (page 137, bottom) gives us a geomet- 
rical way of visualizing vector addition. The algebraic 
way we have already had : when F is ( a , b) and G is (c, d ), 
then F + G is (a + c, b + d). Note that this too appears 
in the figure on page 138. 

Vector addition, of course, plays an important part in 
mechanics. It appears in the combination of forces, 
velocities, and accelerations. Vectors pervade many 
branches of theoretical physics — electromagnetic theory 
for instance. 

If A: is a number, kF is defined as the vector ( ka , kb). 
This is the vector having the same direction as F, but 
its arrow is k times as long. The diagram here shows F 
and 2\F. F is OA. 2\F is OE. 

Dimension of a Space 

So far we have been talking of displacements so much 
to the East and so much to the North. Such displace- 
ments will get us anywhere we want on the ground. 

If we want to rise in the air or bore into the ground, 
we must consider displacements a inches to the East, 
b inches to the North, c inches Up. Starting from a point 


A Concrete Approach to Abstract Algebra 



on my desk, by suitable choice of a, b, c, I can get any- 
where in the room. Above the desk, c will be positive, 
below negative. This displacement we can denote briefly 
by the symbol (a, b, c). 

It is still quite simple to add displacements, by con- 
sidering their combined effect. If I move an object 
2 inches East, 3 inches North, and 4 inches Up, and then 
later I move it a farther 5 inches East, 6 inches North 
and 8 inches Up, the total effect will be 7 inches East, 
9 inches North, and 12 inches Up. In symbols 

displacement A is (2, 3, 4) 

displacement B is (5, 6, 8) 

displacement A + B is (7, 9, 12) 

The numbers in the bracket for A + B are found by 
three ordinary addition sums. More generally, if 

F is ( a , b, c ), 


G is (<7, £, /), 


F T - G is (a + d, b -f- e, c + /)• 

We still have to explain what we mean by kF where 
A; is a number. 

Linear Dependence and Vector Spaces 


If A; is a whole number, for example 3, we expect 3 F 
to mean the same as F + F + F. If 



F is ( a , b, c ), 
F is ( a , b, c), 
F is (a, b, c ), 

then on adding, 

F + F + F is (3a, 3 b, 3c). 

Geometrically, this is the effect of doing the displace- 
ment F three times. 

In the same way, for any whole number n, it seems 
reasonable to say that nF means ( na , nb, nc). 

How about fractions? What shall we understand by 
f F? Suppose this is called X. By X = §F we should nat- 
urally understand that 5X = 3 F. Now this fixes X. If X 
is {u, v, w), 5X is (5w, 5 i», 5 w), and 3Fis (3a, 3b, 3c). This 
means that 

5m = 3a, Sv = 3b, 5 w = 3c. 


m = §a, v = %b, w = f c. 

Hence X is (f a, %b, f c). 


A Concrete Approach to Abstract Algebra 

So the definition that 


kF is ( ka , kb, kc ) 
F is (a, b, c ) 

agrees with our ordinary ideas when £ is a natural num- 
ber or a positive fraction. We shall still use this definition 
when k is negative, like — 3, or irrational, like V: 2. 

Now let P stand for 1 inch to the East, 

Q stand for 1 inch to the North, 

R stand for 1 inch Up. 

Earlier we had the symbol (2, 3, 4) for A. What is the 
corresponding symbol for P? P is 1 inch East, nothing 
North, nothing Up. So P is (1, 0, 0). In the same way, Q 
is (0, 1, 0), and R is (0, 0, 1). 

Now A, being 2 inches East, 3 inches North, and 
4 inches Up, could evidently be gotten by doing the dis- 
placement P twice, then Q three times, and R four times. 
This suggests that A = 2P + 3Q + 4 R. We have reached 
this conclusion by a geometrical argument. We can 
check it algebraically. 

Since P is (1,0,0), IP is (2, 0, 0) 

Since Q is (0, 1,0), 3Q is (0, 3, 0) 

Since R is (0, 0, 1), 4 R is (0, 0, 4). 

Adding, 2P + 3Q + 4 R is (2, 3, 4), which is A. 
Exactly the same argument shows that for F, the dis- 
placement (a, b, c ), 

F = aP + bQ + cR. 

By suitable choice of a, b, c, the vector F can be made 
equal to any displacement. 

Thus, when we are considering displacements of this 
kind, any displacement can be gotten by adding suitable 
multiples of three vectors, P, Q, R. If you think of P, Q, R 

Linear Dependence and Vector Spaces 


as ingredients, every F can be regarded as a mixture, in 
suitable proportions, of these ingredients. 

In our earlier work, when we did not allow displace- 
ments Up, any displacement could be gotten by a suit- 
able mixture of P and Q. Every journey can be achieved 
by traveling East and then North (minus numbers 
accounting for West and South). 

Movement on the ground — or, more generally, in a 
plane — is called movement in two dimensions. Move- 
ment East, North, and Up is called movement in three 
dimensions. In the form “3D,” as applied to motion 
pictures, this latter term has passed into the language of 
the man in the street. 


ent mathematicians use the word “basis” with slightly 
different meanings. In saying that P and Q form a basis 
for vectors in this plane, we shall understand the fol- 

(i) Every vector F can be gotten by a suitable mixture 
of P and Q. 

(ii) Every vector F can be gotten in only one way as 
a mixture of P and Q, 

(iii) Every mixture of P and Q lies in the plane. 


A Concrete Approach to Abstract Algebra 

To illustrate these: P by itself does not form a basis, 
since there are vectors in the plane that cannot be gotten 
by moving Eastward — algebraically, that cannot be rep- 
resented by kP. For example, the vector S with com- 
ponents (1,1) has the direction North-East. Geometri- 
cally it is evident that S cannot be gotten by taking any 
amount of P (that is, by traveling East for any distance) . 

Algebraically, kP is the vector {k, 0), and however we 
choose k we cannot make this into (1, 1). 

Condition (i) thus shows that P alone does not form a 
basis. There are more things that can be done in a plane 
than pure East- West travel. 

Condition (ii) is intended to avoid extravagance. We 
might propose P, Q, and S (S was defined two paragraphs 
earlier) as a basis. P + Q + S would then be the dis- 
placement (1, 0) + (0, 1) + (1, 1) = (2, 2). But (2, 2) 
could equally well be represented as 2S, or as 2P + 2Q, 
or as 3 P + 3 Q — S, or in many other ways. As S is itself 
already a mixture of P and Q, it is wasteful to bring S in. 

Linear Dependence and Vector Spaces 


Condition (i) says, “You must have enough ingredi- 
ents to give you any vector in the plane.” Condition (ii) 
adds, in effect, “But don’t be wasteful.” Some discus- 
sion would be needed to prove that this is just what 
condition (ii) requires for the condition is not worded 
that way. However, if you experiment with choosing a 
basis for the plane, you will get the feeling that this is 
how it works out. 

Condition (iii) is designed to avoid another kind of 
extravagance. P, Q, R might be proposed as a basis for 
displacements on the ground ( P East, Q North, R Up). 
Now it is true that every displacement on the ground 
can be represented as aP + bQ + cR, but of course c 
will always have the value zero. For R takes us out of 
the ground plane. Condition (iii) here requires us not 
to include in our basis any vector that takes us out of 
the ground plane. 

Now, of course, P, Q do not form the only basis for 
this plane. We could equally well take S and T as shown 

here. S, as before, is (1, 1), and is to the North-East; T is 
( — 1, 1), to the North-West. A suitable combination of S 
and T will get us anywhere we want to go. 

Nor is it necessary for the vectors in the basis to be 
perpendicular. A suitable combination of S and P will 
give us any vector F. Of course, negative numbers will 
be needed if F lies in certain parts of the plane. For 

146 A Concrete Approach to Abstract Algebra 

example, F might be taken as Q. How is Q to be gotten 
by mixing S and P? 

S is (1, 1), 

P is (1, 0), 

Q is (0, 1). 

Evidently Q = S — P, which can also be seen geomet- 
rically. The journey S, followed by — 1 to the East (1 to 
the West) brings us 1 to the North, as required for Q. 


Linear Dependence and Vector Spaces 



1. Express S and T as mixtures of P and Q. 

2. Express P and Q as mixtures of S and T. Interpret geomet- 

3. Show that the answer to question 2 could be obtained 
from the answer to question 1, by solving the simultaneous 
equations S = P + Q, T = — P + Q for P and Q. 

4. Which of the following form a basis for vectors in a 
plane? (i) The vectors (0, 1) and (1, 1). (ii) The vectors (1, 2) 
and (2, 1). (iii) The vectors (1, 2) and (2, 4). (iv) The vectors 
(1,5), (2,2), and (4,1). 

5. Does any basis in the plane (i) consist of only one vector? 
(ii) consist of three vectors? 

6. If U is (a, b) and V is (c, d), the vector (p, q) will be ex- 
pressible as xU + yV if x, y satisfy the equations 

p = ax + cy 
q = bx + dy. 

Solve these equations for x and y. 

If U and V form a basis, every vector (p, q) can be expressed 
in the form xU + yV. That is to say, the equations above have 
a solution, whatever p and q. Find the condition a, b, c, d must 
satisfy if this is to be so. 

(If your work seems to show that the equations can always 
be solved, consider the case a = 1, b = 1, c = 2, d = 2. The 
vectors (1,1) and (2, 2) do not form a basis. For these values 
the equations cannot be solved when, for example, p — 1, 

q = 0 .) 

You will have noticed that we had many ways of 
choosing a basis for the ground plane, but that the basis 
always contained two vectors — no more, no less. 

In the same way, if we are allowed to move freely 
(North, East, and Up), we can choose as a basis the three 

148 A Concrete Approach to Abstract Algebra 

vectors P, Q, R. Needless to say, there are many other 
ways of choosing a basis, but however we do it, we al- 
ways find that it consists of three vectors. That is what we 
mean when we say that we live in three dimensions, while 
a creature that can only crawl on the ground is confined 
to two dimensions. 

Now it seems fairly evident to us that a plane is some- 
thing different from a space of three dimensions. How- 
ever, someone might ask us, “How do you know that 
the two things are distinct? Might there not be a space 
such that, by one way of choosing your basis you found 
it had three vectors, and by another way you found a 
basis with two vectors in it?” We certainly do not expect 
such a thing to be possible; but we are now challenged 
to say why. Later a theorem will be given that meets 
this question. 


In the first part of this chapter, we considered packets 
of screws and nails. Since we could only buy a whole 
number of packets, we considered expressions such as 
3 H + 2C in which natural numbers came. Thus, at 
this stage we considered aH + bC with a , b natural 
numbers. This led to the diagram on page 135, in which 
the dots all lay in a certain region — between East and 
North-East from the origin. 

We could extend the network if we allowed a and b 
to take positive or negative whole numbers as values. 
Our dots would then extend in all directions, but of 
course they would not fill the plane ; there would still be 
spaces between them, as at present. 

Now we have a choice in fixing our terminology. Shall 
we say that a vector space must be like a plane — a con- 
tinuous membrane, so to speak — or shall we accept a 

Linear Dependence and Vector Spaces 


network of points as forming a vector space, even if 
there are intervals between the points? 

For a book on algebra, there is no doubt about the 
answer. We have talked about operations + , •, — , -r, 
but nowhere have we said what we mean by “continu- 
ous,” “near to,” or any such term. These are the ideas 
of topology or analysis rather than algebra. Provided a 
structure allows us to add, subtract, and multiply, we 
accept it. 

Accordingly, we shall have a good deal of freedom in 
deciding what coefficients to allow. We might consider 
expressions a U + bV where U and V were given “things” 
(of some kind) and a, b were required to be any one of 
the following: 

(i) Integers (positive or negative), 

(ii) Rational numbers, 

(iii) Real numbers, 

(iv) Complex numbers, 

(v) Numbers modulo 2, 

(vi) Numbers modulo 3, 

(vii) Polynomials in *, with real coefficients. 

This by no means exhausts the possibilities. Some of 
these structures we might be able to realize geometrically, 
others we would be unable to draw. This does not 
matter at all. 

The following example calls attention to a possible 
misunderstanding. Suppose we define U as being 
inches to the East, V as being V 3 inches to the North, 
and require the coefficients a , b to be integers, positive 
or negative. This gives us a perfectly good vector space, 
of the “network” type. You will notice that irrational 
numbers V2, V3 come into the definition of U and V, 
while only integers are allowed for coefficients. It does 
not matter that \^2 and V 3 are not integers. 


A Concrete Approach to Abstract Algebra 

We are not doing anything worse here than when we 
take screws and nails as our basic “things.” Here our 
basic “things” are an arrow pointing Eastward, and an 
arrow pointing Northward. One arrow happens to be 
“s/2 inches long; the other happens to be inches 
long. We then combine a times the first arrow with 
b times the second arrow; a geometrical construction 
shows the meaning of this statement. Thus it is quite 
possible, in the definition of our basic “things” U, V, 
to use numbers that would not be acceptable as coeffi- 
cients a , b. 

In our hardware example, we used natural numbers 
as coefficients. This was quite useful as an introduction, 
to show how expressions like 2 H + 3 C came about. 
However, the natural numbers suffer from the defect 
that you cannot always subtract. You cannot take away 
5 screws and 8 nails from 2 screws and 3 nails. In a vector 
space (as this term is used by the majority of mathe- 
maticians today), it is required that subtraction be al- 
ways possible. That is, if L and M are two vectors, there 
must always be a vector X such that L + X = M, 
and X is called M — L. That is why the natural num- 
bers are not on the list of suggestions for coefficients. 
The set of “possible purchases,” mentioned at the begin- 
ning of this chapter, thus does not constitute a vector 
space. If we extended the definition to include negative 
purchases — a customer returning several packets — then 
it would constitute a vector space over the integers, 
with H and C as a basis. 

Specification of a Vector Space 

We have now reached the stage where we can specify 
what we mean by a vector space. A vector space involves 

Linear Dependence and Vector Spaces 


two sets of objects, rather intimately mixed. I shall call 

(i) Things, 

(ii) Numbers. 

The numbers belong to a certain specified class. By 
the use of the word “number,” I do not imply any 
connection with, for instance, counting. By number I 
just mean something that is going to be used, later on, 
as a coefficient like a, b above. It must be possible to 
add, subtract, and multiply numbers. To be more pre- 
cise, I suppose the numbers to obey axioms (7) through 
(P), (77), and (72). (In our main applications, the num- 
bers will be assumed to obey all the field axioms. Stu- 
dents should be careful to note the point in the develop- 
ment of the theory after which the results are true 
ONLY for fields.) 

The set from which the numbers are drawn will be 
denoted by K. 

There is no restriction on the kind of object that can 
play the role of a Thing. In any particular example, we 
of course specify what is allowed as a Thing. Things 
may be physical objects, ideas in the mind, marks on 
paper, sounds of words, actions, operations. A Thing 
can be, and usually will be, of a compound nature. Thus 
we regard “3 screws and 2 nails” or “5 inches East and 
6 inches South” or “3x 2 — 5 xy + 6 y 2 ” as being a Thing. 
The numbers of K may very well enter into the specifi- 
cation of a Thing, as for example if we take 

K : all real numbers. 

Things: all displacements in the ground plane. 

The real number 3^2, for example, enters into the 
specification of “3^2 inches to the North-East.” 

The Things form a vector space of n dimensions over 
the numbers K if the following statements hold. 


A Concrete Approach to Abstract Algebra 

(I) Every Thing can be given a label (ai, a 2 , • • • , a n ) 
where < 21 , a 2 , • • • , a„ are numbers from K. No Thing 
has two different labels. 

(II) To every possible label there corresponds one, and 
only one, Thing. 

(III) Corresponding to any two Things, A and B, there 
is defined a third Thing, C, called the sum of A 
and B. If A has the label (ax, • • • , a n ), B the label 
( bi , • • • , b n ), C the label (ri, • • • , c n ), then 

ci = a\ + b\ 
c 2 = 02 + b 2 

Cn — CLn ~ f" b n . 

(IV) Corresponding to any Thing A and any number k, 
there is defined a Thing D, called k times A. If 
A has the label (ax, • • • , a„), D has the label 

(Jc&iy * * * , kdn) . 

Some examples of vector spaces will help to make 
these conditions clear. 

Example 7. 

K : the real numbers. 

Things: the displacements in the ground plane. 

We use (a, b) as a label for “a inches East and b inches 
North.” This provides a label for every possible displace- 
ment. There is a displacement corresponding to every 
possible label. We never get two different labels for the 
same displacement, nor two different displacements for 
the same label. 

We have already seen how statements (III) and (IV) 
apply. We have here a space of 2 dimensions over the 
real numbers. 

Example 2. 

K : the integers. 

Things: all expressions ax + by + cz with whole 
number coefficients a, b, c. 

Linear Dependence and Vector Spaces 


Addition of expressions and multiplication of expres- 
sions by numbers are defined as in elementary algebra. 

Thus, for example, the sum of 2x + 3y + 4z and 
5x + 6y + 7z is lx + 9y + 1 lz; and 5 times 2x + 3y + 4 z 
is 10x + 1 5jy + 20 z. 

We choose the label (a, b, c ) for the expression ax + 
by + cz. Thus (2, 3, 4) is the label for 2x -f- 3y + 4 z. No 
other expression gets this label. No expression has two 
different labels. 

Does statement (III) apply? 

We take an example: 

2x + 3y + 4 z has label (2, 3, 4), 

5x + 6y + lz has label (5, 6, 7). 

If statement (III) holds, the sum should have label 
(7, 9, 11). 

In fact, it does so. One can show, by simple algebra, 
that statement (III) applies generally. 

Does statement (IV) hold? Again, we take an exam- 
ple. The expression 2x + 3y + 4 z has label (2, 3, 4). If 
statement (IV) holds, 5 times the expression should have 
label (10, 15, 20). It does; and we can show that this 
always happens. 

We have here a vector space of 3 dimensions over the 
integers. Every expression ax + by + cz is a mixture of 
x, y , and z. x , y, z form a basis. You may notice that x 
has the label (1, 0, 0), y the label (0, 1, 0); z the label 
( 0 , 0 , 1 ). 

Example 3. 

K : the real numbers. 

Things : all quadratics ax 2 + bx + c with real a, b , c. 

This is a vector space of 3 dimensions over the real 
numbers, x 2 , x, 1 form a basis, ax 2 + be + c receives the 


A Concrete Approach to Abstract Algebra 

label (a, b , c ). * 2 has the label (1, 0, 0), x the label (0, 1,0), 
1 the label (0, 0, 1). 

Example 4. 

K: the rational numbers. 

Things: all numbers a + by/ 2 with a, b rational. 

We assign the label (a, b) to a + by/ 2. Thus 5 + 7 a/ 2 
would be labeled (5, 7). It is clear that only one Thing, 
a + bV 2, corresponds to the label ( a , b). But what about 
the other way? Can two different labels give the same 
Thing? Suppose they could ; imagine that {a, b) and ( c , d) 
are two labels for the same Thing. This means that 

a + bV 2 = c + dV2,_ 

a — c — (d — b)y/ 2. 

If d — b is not zero, we can divide by it. Then 
(a - c)/(d - b) = V2. 

This gives V2 as a fraction, a rational number. But 
this has been proved impossible. So d — b must be zero. 
It follows that a — c is zero also. Hence d = b, a = c. 
So the two labels are the same. 

Statements (III) and (IV) can be checked without 
difficulty. We thus have a vector space of 2 dimensions 
over the rationals: 1 has the label (1, 0), y/ 2 the label 
(0, 1). Together 1 and V 2 form a basis. 

Example 5. 

K: 0, 1 modulo 2 
Things: 0, 1, M, M + 1. 

The Things here form GF{ 2 2 ). All the Things are ob- 
tained by letting a and b run through the values 0, 1 in 
the expression a + bM. We have a vector space of 2 di- 
mensions over K. 

Linear Dependence and Vector Spaces 


1 has the label (1, 0), and M the label (0, 1). 
1 , M form a basis. 

Example 6. 

K: the integers. 

Things: a + %b, (a, b integers). 

This has perhaps the appearance of a space of 2 dimen- 
sions with (a, b ) as the label for a + \b. But then (3, 0), 
(2, 2), (1, 4), (0, 6), and many others would all label the 
same thing. 

In fact, of course, whatever integers you choose for 
a, b, the value of a + \b will lie in the set 

These are all of the form \m where m is an integer. 
We have a space of 1 dimension over the integers. Basis, 
the single Thing, f . ( m ) is the label for \m. 

Contrast example 4. 

Example 7. 

K: the integers. 

Things: all numbers of the form a h 
where a, b are integers. 

Sum and product: as in arithmetic. 

This, of course, is not a vector space over the integers. 

If b is positive, a b is an integer. If b is negative, say 
b = —m, then a b = a~ m = \/a m and a b is the reciprocal 
of an integer. 

Now 3 _1 = 1/3 and 5" 1 = 1/5, but 3” 1 + 5" 1 = 1/3 + 
1/5 = 8/1 5, which is neither an integer nor the recip- 
rocal of an integer. So A = 3 _1 , B — 5 -1 are Things, 
but A + B is not a Thing. 

There is another way in which this structure fails to 
be a vector space that is worth noting. Suppose we de- 
cided to take {a, b) as a label for a b . 


A Concrete Approach to Abstract Algebra 

a h gets the label {a, b ), 
c d gets the label (c, d). 

If statement (III) holds, a b + c d ought to get the label 
(a + Cy b + d), but a b + c d is not equal to {a + c) {b+d \ 

Chapter 8 

Algebraic Calculations 
with Vectors 

We saw in Example 2 (page 153) that linear expressions 
of the form ax + by + cz constituted a vector space of 
3 dimensions. Thus a vector space is no unfamiliar thing ; 
if you have handled linear expressions in school algebra, 
you already know a vector space and how it behaves. 
But, more than this, vector spaces have very little indi- 
viduality; they differ of course in dimension, but this is 
not a very serious difference — if you can correctly per- 
form algebraic calculations on expressions of the type 
ax + by + cz, you will hardly find difficulties with the 
type ax + by or the type ax + by + cz + dt. Vector 
spaces also differ in the numbers K they employ: it 
makes some difference whether a, b, c are integers, or 
rational, or real, or numbers modulo 5 — but again, not 
very much difference; algebraic calculations with all of 
these have much in common. Working with vector spaces 
is accordingly much the same thing as working with linear 
expressions in elementary algebra. 

You may have noticed that all the equations we have 
had connecting vectors have been linear equations, for 
example : 


A Concrete Approach to Abstract Algebra 

F = aP+bQ+cR (page 142) 

Q — S — P (page 146) 

Linear expressions are very simple to handle. Geomet- 
rical questions, which otherwise might be difficult, can 
sometimes be expressed in the language of vectors, and 
thus converted into simple algebraic problems. 

Example. Three wires radiate from a point 0. The 
first wire is horizontal and points due East. The second 
wire makes an angle of 45° with the horizontal and goes 
North from 0. The third wire goes in a Northeasterly 
direction, and rises 1 foot for every V2 feet in a hori- 
zontal direction. Do the three wires lie in a plane or not? 

Solution. We use the symbol {a, b , c) as on page 140. 
The first wire evidently has the same direction as the 
vector (1, 0, 0) = A, say. The second wire has the direc- 
tion of the vector (0, 1, 1) = B , say. The third wire has 
the direction of the vector (1,1,1) = C, say. (A diagram 
or model will make these statements clear.) 

Does C lie in the plane of A and B? To answer this 
question we must translate into the language of vectors 
“the plane of A and B.” This is not hard to do. 

If we go any distance in the direction of A, and then 


Algebraic Calculations with Vectors 


any distance in the direction of B , the total displacement 
is one lying in the plane of A and B. 

The vector corresponding to “any distance in the 
direction of A” is represented by sA, for some number s. 

The vector corresponding to “any distance in the 
direction of i?” is represented by tB, for some number t. 

The combined effect of these displacements is gotten 
by addition (see page 139). It is thus sA + tB. 

Hence any vector in the plane of A and B is of the form 
sA ~f- tB. 

Our question becomes, can we find numbers s and t 
for which C = sA + tB? 

This question can in fact be answered immediately. 
We have 

A = (1, 0, 0). 

B = (0, 1, 1), 

A + B = (1, 1, 1) = C. 

So s = 1, / = 1 does the trick. 

If we did not spot this, we should have to proceed as 

A = (1, 0, 0). sA = (. r, 0, 0). 

B = (0, 1, 1). /. tB = (0, t, t). 

sA + tB = (.r, t, t). 

We want, if possible, to make this equal to (1, 1, 1). 
s = 1 , t = 1 does this. 

So the three wires do lie in one plane. 

The connection of this work with linear expressions 
appears most clearly if we use the three vectors P , Q, R 
as defined on page 142. (The symbols A, B we are now us- 
ing have of course no connection with the symbols A, B 
occurring on pages 139-142.) Then 

A = P 

B= Q + R 
C = P+ Q + R 


A Concrete Approach to Abstract Algebra 

and the question is, can we find numbers s, t so that 
C = sA + tB ? To see that s = 1, t = 1 provides a 
solution requires no knowledge beyond very elementary 

Someone who was told to handle vector expressions 
aP + bQ + cR “just like ax + by + cz in elementary 
algebra” would probably find this quite a sufficient 
guide for practical calculation. However, this prescrip- 
tion is somewhat vague, and it will be as well to list the 
properties we have in mind. 

For definiteness, we shall take n = 4, so that our 
vectors will have labels (01, a 2 , <33, « 4 ). But the arguments 
apply equally well whatever natural number is chosen 
for n. 

The numbers, shown by small letters, ai, b 2 , k, and so 
on, are supposed to obey axioms (7) through (9) and 
(77). The vectors, or “Things,” are subject to statements 
(I)-(IV) of page 152. 

Denote by Pi the vector with the label (1, 0, 0, 0), 

Denote by P 2 the vector with the label (0, 1, 0, 0), 

Denote by P3 the vector with the label (0, 0, 1, 0), 

Denote by P 4 the vector with the label (0, 0, 0, 1). 

It requires no ingenuity to verify the results listed 
below. Several of these are left as exercises. 

(V.l) Corresponding to any two vectors, A and B, 
there is a vector C, called A + B. This is given 
in statement (III), page 152. 

(V.2) A + B = B + A. This follows from the latter 
part of statement (III), and from axiom (3) for 
numbers a T , b r . 

(V.3) A + (P + C) = (A + B) + C. 

(V.4) There is a vector 0 such that A + 0 = A for 
any vector A. 

Algebraic Calculations with Vectors 


(V.5) The equation A + X — B has a unique solution 
X, whatever the vectors A and B. X is called 
B - A. 

(V.6) For any number k and any vector A, the product 
k'A is uniquely defined. This is the earlier part 
of statement (IV). 

(V.7) For any number k and any vectors A, B , 
k-(A + B) = (k-A) + ( k-B ). 

(V.8) For any numbers a, b and for any vector A, 

(a + b)- A = (a- A) + (/ b-A ). 

(V.9) For any numbers a, b and for any vector A, 

(< ab ) • A = a -(b-A) 

(V.10) Every vector A is expressible in the form 

a\P\ + & 2 P 2 + Q 3 P 3 + aiPi. 

(This is for n — 4. In general, aiP x + • • • + a n P n of 
course replaces the expression here.) 


1. Prove statements (V.3, V.4, V.5, V.7, V.8, V.9, V.10) 
above, the statements (I)-(IV) of page 152, and the axioms 
( 1)-(9 ) and (77) for the numbers being given as known. 

2. Prove from the statements (V.l)-(V.IO), that for the vec- 
tor 0 and for any number k , we have k-0 = 0. 

3. Prove that, if 1 is the unit of the number system, for any 
vector A, we have 1-A = A. (Begin with V.10. Use V.7 and 

4. Prove that, 0 denoting the zero of the number system, and 
A being any vector, we have 0-A = O. 

5. Prove that, if —A is used as an abbreviation for 0 — A, 
then — A = ( — 1)-^4. 


A Concrete Approach to Abstract Algebra 

6. What statements from (V.l)-(V.IO) are used in obtain- 
ing the following results? (i) 2 • (34 + 4 B) + 5 • (34 + 45) = 
7- (3 A + 45). (ii) 3 • (5^4 + 65) = 154 + 185. 

The above examples show how closely calculations 
with vectors resemble elementary algebra; in fact, as 
was mentioned earlier, linear expressions ax + by + 
cz + dt actually constitute a vector space. 

Thus, in order to work with vectors, no new formula, 
no new procedure needs to be memorized. 

Statements (I) through (IV) of page 1 52 are needed 
if one is in doubt whether or not some particular struc- 
ture is a vector space. 

Statements (V.l) through (V.10) resemble several of 
the axioms for a field (page 27). Of course, they contain 
nothing resembling field axiom (70), that provides for 
division. These statements, (V.l) through (V.10), can 
always be appealed to, if our right to perform some for- 
mal algebraic manipulation with vectors is challenged, 
or if we are in doubt whether or not a particular step 
is justified. In some treatments of vector spaces, state- 
ments (V.l) through (V.10) form the starting point, 
rather than statements (I) through (IV). 

Linear Independence 

Suppose, in elementary algebra, we consider all ex- 
pressions of the form au + bv ~h cw, where a, b, c are 
integers and 

« = x+ y + z + ?, 
v = x + 2y + 3z + At, 
w = x + 4y + 7z + 10£. 

At first sight, you might think these expressions con- 
stituted a vector space of 3 dimensions, basis u, v, w, with 
(a, b, c) as the label for the expression au + bv + cw. 

Algebraic Calculations with Vectors 


However, consider the two cases 

(i) a = 1 b = 5 c = 3, 

(ii) a = 3 b = 2 c — 4. 

We find 

u + 5v + 3 w = 9x + 23y + 37z -f- 5l£, 

3m + 2m + 4za = 9* -f 23jy + 37z + 51 1 . 

That is to say, the labels (1, 5, 3) and (3, 2, 4) both 
describe the same Thing. This, however, is contrary to 
statement (I). Since 

u + 5v + 3w = 3u T" 2v + 4 w, 
it follows that 

2u — 3v + w = 0. (1) 

In fact, w = — 2u + 3v, so any expression au + bv + cw 
could be written au + bv + c( — 2u + 3v). This equals 
{a — 2c) u + (b + 3 c)v, which is a mixture of u and v 
alone. For example, u + Sv + 3w and 3m + 2v + 4 w 
above are both expressible as — 5m + 14m. 

Thus it now appears that we have a vector space of 
only 2 dimensions, with u, v as a basis. But perhaps this 
is not yet the end of the story; perhaps m, v can be re- 
placed by something even simpler. We now examine 
this possibility. 

Suppose we use (a, b) as the label for the expression 
au + bv. Can one expression receive two labels? Suppose 
it could ; suppose ( p , q) and (r, s) represented the same ex- 
pression, that is, 

pu + qv = ru + sv. (2) 


We write 

(p — r)u + (q — s)v = 0. 
f = p - r 


A Concrete Approach to Abstract Algebra 


u = x-\-y-\-z-\-t, 
v = x + 2y + 3z + At. 

For fu + gv to be zero, we must have (as is seen by con- 
sidering the coefficients of x, y, z, t ) 

f + g — 0 
/+ 2 £ = 0 
/+ 3 ^ = 0 
f+4g = 0 

( 5 ) 

The only solution of these simultaneous equations is 
/ = 0, g = 0. By equations (3), this means that p — r, 
q = s. And this means that (p, q) and (r, s) are one and 
the same label. 

So it is impossible for one expression to receive two 
distinct labels. With u, v as a basis, all the requirements 
of statements (I)-(IV) are met, and our expressions 
thus form a vector space of 2 dimensions. 

In this investigation, one or two ideas have occurred 
that are of value for the theory generally, so that it is 
worth while to introduce names for them. 

In testing whether u, v formed a basis for a space of 
2 dimensions, we considered the mixture fu + gv. We 
found that this was zero only when both / and g were 
zero. The only way of mixing u and v so as to obtain 
zero was to take none of either. 

We then say that u, v are linearly independent. 

Generally, we say that n vectors u h w 2j • • • , u n are lin- 
early independent if the only way to make the mixture 
ciui + C 2 U 2 + • • • + c n u n zero is to make c\ — 0, r 2 = 0, 

* ' ' ) Cn ~ 0. 

On the other hand, we had above u, v, w for which the 
mixture 2u — 3v + w was zero. This is a genuine mix- 
ture; it is not just “none of u , none of v, none of 
Accordingly, we say that u, v, w are linearly dependent. 

Algebraic Calculations with Vectors 


definition. If n vectors , «i, M2, • • • , M ft satisfy a linear 

C\U\ + C 2 U 2 + * * • + C n U n = 0, 
other than the trivial equation, 

Omi -j- OM 2 “t~ * • • 4~ 0u n = 0, 

which is satisfied by any set of n vectors, we call the vectors 
“ linearly dependent .” 

You will notice that “linearly independent” is another 
way of saying “not linearly dependent.” A set of vectors, 
Mi, • • • , M n , must be one or the other. 

Geometrical Interpretation 

Geometrical examples of vector fields are helpful for 
illustrating the meaning of linear independence. 

For instance, on page 142 we met vectors P,Q,R which 
represented an inch to the East, an inch to the North, 
and an inch Up, respectively. Are these linearly inde- 
pendent or not? That is, can we find a genuine mixture 
of them that is zero? If we take cfP + c 2 Q + c 3 R, this 
represents the displacement c\ inches to the East, c 2 
inches to the North, c 3 inches Upwards. It is clear that 
this gives the zero displacement only if ci — c 2 = c s = 0. 
That is, only the trivial mixture gives zero. Hence 
P, Q, R are linearly independent. 

If, on the other hand, we take the vectors P, Q, S of 
page 144, we have P + Q — S = 0. The vectors are 
linearly dependent. 

Quite generally, if U, V, W are three vectors in 
geometrical space of 3 dimensions, they are linearly 
dependent if they satisfy a non-trivial equation. 

aU+bV +cW = 0. 

Then the journey a U (which is in the same direction as 


A Concrete Approach to Abstract Algebra 

U), followed by the journey bV and the journey cW, 
brings us back to the point where we started. Evidently 
this means that U, V, W lie in a plane. 


By a model or diagram make clear to yourself how the vectors 
listed below are situated in space. Hence determine whether 
or not each set is linearly independent. Check your conclusions 

1 . A = (1,1,0), 

B = (-1, 1,0), 

C = (0, 0, 1). 

2. A = (2, 3, 0), 

B = (10, 1, 0), 

C = (1, 1, 0). 

3. A = (0, 1, -1), 

B = (-1,0, 1), 

c = ( 1 , - 1 , 0 ). 

4. A = (0, 1, 1), 

B = (1, 0, 1), 

c = ( 1 , 1 , 0 ). 

Chapter 9 

Vectors Over a Field 

The processes we are soon to use involve division 
by a number. From now on, we assume that our number system 
forms a field. 

On page 159 we saw that the vectors of a plane were 
all of the form sA + tB. 

In the same way, all the vectors that lie in a line are 
of the form sA. Those that fill a 3-dimensional space 
are of the form sA + tB + pC. 

It is thus natural to study mixtures of a given set of 
vectors. We may in this way obtain vector spaces that 
do not correspond to our geometrical ideas derived from 
the physical world. By considering mixtures of linear 
expressions in ten variables we could construct, alge- 
braically, vector spaces of any number up to 10 dimen- 
sions. Needless to say, we cannot visualize such spaces 
geometrically, though we may be helped in dealing with 
them by analogies drawn from our physical experience 
of three dimensions. Vectors in such spaces are not with- 
out practical interest. In intelligence testing, for instance, 
a candidate might be subjected to ten examinations. If 
candidate A scored (ai, ai, • • • , flio) in these examina- 
tions, his performance would specify a vector in 10 
dimensions. If candidate B scored bi, • • • , bi Q and candi- 


A Concrete Approach to Abstract Algebra 

date C scored a, • • • , do, one might find, perhaps, that in 
each test C’s score was the sum of 4’s and B’ s. This we 
could write as C = A B. Or C’s marks might be, 
in each examination, the average of T’s and B’s. Or 
again, if A scored full marks on test 1 , and almost nothing 
on the others while B scored high marks on test 10 and 
nothing on the others, we could say that A’ s and B s 
abilities lay in different directions. All of this goes quite 
naturally in the language of vectors, and in fact vectors 
are used considerably in the theories of intelligence. We 
are, then, concerned with mixtures of vectors, whether 
these can be illustrated in physical terms or not. 

We saw on page 163 that a mixture of three vectors 
might sometimes produce a vector space of only 2 dimen- 
sions. It is desirable to have a systematic way of deter- 
mining the dimensions of the space generated by a col- 
lection of vectors. (The space formed by all possible 
mixtures of a given set of vectors is said to be generated 
by these vectors.) This is quite simple to do. The process 
depends on the two following principles. 
principle i. The space generated by a set of vectors is 
unaltered if any vector P is replaced by kP, where k is any 
non-zero number. 

principle ii. The space generated by a set of vectors is 
unaltered if any two vectors P, Q of the set are replaced by 
P, Q — sP where s is any number whatever. 

Principle I, applied to a set of three vectors A, B, C, 
states that kA, B, C generate the same space, that is, 
any mixture of kA , B, C is a mixture of A, B, C, and con- 
versely. That a mixture of kA, B, C is a mixture of 
A, B, C is almost obvious. That a mixture of A, B, C 
can always be expressed as a mixture of kA, B, C appears 
from the equation 

aA + bB + cC = ( a/k ) • (kA) + bB + cC. 

Vectors Over a Field 


No new idea is needed to prove this principle for any 
number of vectors. 

The way in which principle II is proved is also clear 
from the particular case of three vectors. We want to 
show that P, Q, R and P, Q - sP, R generate the same 
space. We write 

U = P, 

V = Q — sP, 

W= R. 

It is to be shown that any mixture of U, V, IT is a mix- 
ture of P, Q, R, and conversely. That a mixture of U, 
V, IT is a mixture of P, Q, R is obvious. But since 

P = U, 

Q= V+sU, 

R = W, 

it is also obvious that any mixture of P, Q, R is expressible 
as a mixture of U, V, W. 

The proof for any number of vectors follows exactly 
the same lines. 

Principle II can be applied repeatedly. For instance, 
by applying it three times, one can see that (P, Q, R, S) 
and (P, Q — aP, R — bP, S — cP ) generate the same 

Great caution is needed to avoid jumping to unwar- 
ranted conclusions. One might think that P, Q, R and 
P — Q, Q — R, R — P would generate the same space. 
This, however, cannot be derived by any number of 
applications of principle II, and is in fact untrue; con- 
sider for example the case when P, Q, R are all equal. 

Conditions for a Space of n Dimensions 

Suppose now we take any n vectors P u P 2 , • • • , P n , and 
consider all the vectors of the form afPi + a 2 P 2 + • • • + 


A Concrete Approach to Abstract Algebra 

a n P n • Do these form a space of n dimensions? Conditions 
(II), (III), (IV) of page 152 are automatically satisfied, 
(fli, <22, ‘ • , a n ) being regarded as the label for aiPi + 
G2P2 + • • • + a n P n . The only doubt is in regard to the 
latter sentence in condition (I). As we saw in the ex- 
ample on page 163 , this can fail: two different labels may 
correspond to the same thing. 

What is the condition for this failure? If (ai, • • • , a n ) 
and (b h • • • , b n ) are two labels for the same vector, this 
means that 

a\Pi + aiPi + • • • + a n Pn — b\P\ + 62P2 + • • • + b n P n , 

(«1 — 61 ) -Px + ((22 — 62) 'P2 + • * • + ( a n — bn) • Pn = 0 . 

Now (ai, • • • , a n ) is not the same as {b\, • • • , b n )- This 
means that at least one of the coefficients above is not 
zero. Hence, Pi, P 2 , • • • , P n are linearly dependent, for 
they satisfy a non-trivial equation (see definition, page 
165 ). 

Accordingly, if Pi, • • • , P„ are linearly independent, 
the failure of condition (I) cannot occur. All the condi- 
tions (I)-(IV) are then satisfied, and we have the fol- 
lowing theorem. 

theorem. If Pi, •••, P n are n linearly independent vec- 
tors, they generate a space of n dimensions; that is, the totality 
of vectors of the form a\Pi a n P n constitute a space of 

n dimensions. 

A Standard Form 

We now consider two examples in which it is immedi- 
ately evident that a set of vectors are linearly inde- 
pendent. These examples are significant. It will be 
shown later that any set of vectors can be reduced to 
this form by means of principles I and II. 

Vectors Over a Field 


Example. What is the dimension of the space gen- 
erated by P, Q, R, S here specified? 

P = x + 4y — I7z + 3 1 — 2 u, 

Q = v + 5 z — 81* + 39m, 

R — z -f- At “f" 111 m, 

S = t - 13 M. 

By the theorem above, we shall be sure that the space 
is of 4 dimensions, if we can prove P, Q, R, S linearly 

Suppose P, Q, R, S satisfy an equation 
aP + bQ + cR + dS = 0. 

Consider the coefficient of *. x appears only once on the 
left-hand side in the term ax. Hence, a — 0. Accordingly, 

bQ + cR + dS = 0. 

Now y appears only in the term by. Hence, b = 0. We 
substitute b = 0 in the equation, and go on to consider 
the coefficient of z. This shows c = 0. Finally, from the 
coefficient of t, we see d = 0. 

Hence, P , Q, R, S satisfy only the trivial equation 
with all coefficients zero. Hence, by definition, they are 
linearly independent. So the space generated by P, Q, 
R, S is of 4 dimensions. 

You will notice in this example that the coefficients 
(4, -17, 3, -2 inP; 5, -81, 39 in Q; 4, 111 in R; -13 
in S) are never mentioned in the proof. The proof de- 
pends only on the fact that the non-zero coefficients form 
the following pattern. 

j * * * * 

j * * * 

j * * 

1 * 

It is also possible to have examples in which the pattern 

172 A Concrete Approach to Abstract Algebra 

shows steps rather than a diagonal edge, for example 
the following. 

-j ******* 

^ * * * * 

1 * * 


Question: Show that a space of 4 dimensions is gener- 
ated by the vectors P, Q, R, S where 

P = x-\-y-{-z + t J r u + v + w + s, 

Q = t "-f- 2 u 2v 2 w -f- 2s, 

R — v + 3w + 3s, 

S = s. 

In example 2 on page 153, we saw that the label 
(a, b, c ) can be attached to the expression ax + by + cz. 
Our two examples above could equally well be expressed 
by means of such labels. 

Thus, the first example would show that a space of 
4 dimensions is generated by the vectors 

(1, 4, -17, 3, -2), 

(0, 1, 5, -81, 39), 

(0, 0, 1, 4, 111), 

(0, 0, 0, 1, -13). 

It is good to be able to follow the argument in either 
form of notation. It is, of course, essentially the same 
argument in either notation. One can pass from a ques- 
tion stated in bracket notation — that is, with signs such 
as (a, b, • • • , k ) — to the linear expression notation, by the 
process used on page 159. 

In the two examples we have just had, we may say 
that the system P, Q, R, S is in standard form. In the first 
example, P contains the symbol x, which is not in any 
later expression Q, R, S. Q contains y, which is not in 
R, S. R contains z, which is not in S. S contains t. (S is 

Vectors Over a Field 


not followed by any expression. So all we require of S 
here is that it should not be identically zero. If it were 
identically zero, we should simply omit it, since it would 
make no contribution to the space.) 

In the second example, P contains x, which is not in 
any later expression Q, R, S. Q contains t, not in R, S. 
R contains v, not in S. And S contains s. 

These remarks should convey what is meant by “stand- 
ard form.” 


1. Define formally when n linear expressions X\, • • • , X n in 
m symbols x\, • • • , x m are in standard form. 

2. Make clear to yourself how you would recognize that n 
vectors expressed in bracket notation were in standard form. 
(The verbal definition of such a property is often difficult and 
tedious. The idea of standard form can certainly be grasped 
by students who are unable to formulate it verbally. For our 
purposes, the ability to recognize standard form is important j 
the verbalization, less so.) 

Reduction to Standard Form 

Our principles I and II allow us to make chains of, 
arguments such as the following : 

Space generated by A, B, C 

= space generated by A, B — 2A, C by (II) 

= space generated by A, B — 2 A, C — 3 A by (II). 

We thus move from the set of vectors A, B, C to the set 
A, B — 2A , C to the set A, B — 2 A, C — 3 A. These sets 
of vectors are of course distinct. We could not write 
that these sets were equal. But they have the property 
that they generate the same space. 


A Concrete Approach to Abstract Algebra 

Accordingly, when we speak of reducing a set of vec- 
tors to standard form, this is a brief way of saying that 
we are looking for another set, a simpler set, that generate 
the same space as the original set. For example, if 

A = x + 5y + 12 z, 

B = 2x + Uy + 33z, 

C = 3x -f 9y + z, 

the argument above leads us to 

A = x + 5 y + 12 z 
D = B - 2A = y+9z 

E = C - 3A = -6y - 35 z. 

A, D, E generate the same space as A, B, C. Now 
A, D, E are not yet a standard set. But they are nearer 
to standard form than A, B, C since x occurs in A but not 
in B , C. 

To complete the reduction to standard form, we 
apply principle II again, and replace (A, D, E) by 
(A, D, E + 6 D). This yields 

A = x + 5y + 12z, 

D = y + 9z, 

F = E + 6D = 19 z. 

Finally, by principle I, the expressions A, D, F can be 
replaced by A, D, F/ 19, and standard form is thus 

A = x + 5y + 12z, 

D = y + 9z, 

G = 2 . 

A, B, C generate the same space as A, D, G \ evidently 
this is a space of 3 dimensions. 

Question: There is an even simpler set of three vectors 
that generates this space. What is it? 

The procedure followed in reduction to standard form 

Vectors Over a Field 


should be clear. First we select a symbol (x) that actually 
occurs in A. By “actually occurs 5 ’ I mean that its coeffi- 
cient is not zero. Suppose, then, that A contains the 
term ax. By principle I, we can replace A by (1/a) -A. 
We thus have an expression 

P = (\/a)‘A = x + ••• . 

If B contains bx, we replace B by B — bP, which does 
not contain x. In the same way, we can replace C, con- 
taining cx, by C — cP, which contains no x. 

Thus we arrive at a set P, Q, R, S, • • • , in which P 
alone contains x. We set P on one side. We can now 
forget about x, since Q, R, S, • • • , do not contain it. We 
choose a symbol y that actually occurs in Q, and follow 
the same kind of procedure to replace R, S, • • ■ , by ex- 
pressions free from y. We continue until a standard form 
is reached. 

Example. Reduce A, B, C to standard form, where 

A = x y z 2t + m, 

B = 2x + 5y + Sz + It + 5m, 

C — x + 2y + 4z + 2t — u. 

A contains x. We subtract from B and C multiples of A 

so chosen that x disappears. We thus reach 

^4 = x + y T" z -j- 2t -j- u, 

D — B — 2A = 2>y + 6z + 3t + 3m, 

E = C — A = + — 2m. 

The coefficient of y in D is 3. So we use principle I 
and consider ^Z). (It does not matter if such a step intro- 
duces fractions. We are working with the field of rationals. 
Actually, no fractions occur in this example.) 

A = x -}- y H - z T 2l -f* m, 

F = \D = y + 2z + t + m, 

E = y + 3z — 2m. 


A Concrete Approach to Abstract Algebra 

Finally, we have 

A = x + y -f- z It -\~ u, 

F = y + 2z + t + w, 

G = E — F = z — t — 3u, 

which is in standard form. 

Example. Reduce to standard form the expressions 
u, v, w of page 162. These expressions, you may remem- 
ber, generated a space of only 2 dimensions. We shall 
see how this appears in the work. 

u = x+y-\-z+ t, 
v — x + 2y + 3z + At, 
w = x + Ay + Iz + 10?. 

x occurs in the top row. We use principle II to get 
rid of it in the other rows. 

u — x + y + z + t, 
p = v — u = y + 2z + 3/, 

q = w — u = 3y + 6z + 9t. 

y occurs in p. We get rid of y in the third row by form- 
ing q — 'ip. This, however, is Oy + Oz + 0;, the vector 
zero. It makes no contribution to the space generated, 
and is accordingly omitted. Thus the standard form is 

u — x + y + z + t, 
p — 4" 2 z -f- 3t. 

The two vectors u, p generate the same space as the 
original three vectors u , v, w. 

The fact that q — 3p is zero shows that u, v, w are 
linearly dependent. For q = w — u, and p — v — u. 
Substituting for p and q we obtain 

0 = q — 3p = (w — u) — 3(v — u), 

= 2u — 3v + w, 

as in equation (1) of page 163. 

Vectors Over a Field 

1 77 


Reduce the following systems to standard form. Whenever a 
zero vector appears in the work, verify that the corresponding 
vector in the original system is a mixture of the vectors which 
precede it. (In our last example, the third vector in the process 
of reduction is zero; this corresponds to the fact that the third 
vector, w, of the original system u, v, w can be expressed as 
— 2 a + 3v.) Find the relationship in each system. 

1. A = x+ y + z, 

B = x + 3y + 3z + 2t, 

C = 2x + 3y + 6z + At + 3t/. 

2. A = x + 2y + 3 z + At, 

B = 5x + 6y + 7z + 8*, 

C = 9x + lOy + llz + 12#, 

D = 13x + 14 y + 15z + 16*. 

3. A = x — 2y + z, 

5 = * + jy — 2z, 

C = — 2x 4“ yd - z . 

4. ^4 = x — y, 

B = y - z, 

C = —x + z. 

5. ^4 = x + y, 

R = y + z, 

C = x +z. 

We have had several examples in which the appear- 
ance of a zero vector in the reduction process indicated 
that a vector of the original system was a mixture of the 
preceding vectors. We have not, however, proved that 
this must be so. 

The proof requires only one point to be established. 
In our worked example on page 176, we had q — 3p = 0. 


A Concrete Approach to Abstract Algebra 

p and q were linear expressions in the original symbols 
u, v , w. It seems evident that substitution for p and q in 
terms of u, v, w must give a relation au + bv + cw = 0. 
There is, however, one loophole that must be filled; it 
might be that a , b, c all happen to be zero. In that event 
we have not shown u , v, w to be linearly dependent. 

In actual fact q — 3p = 2u — 3v w and the co- 
efficient of w is 1. This is no accident. The process we 
have described is such that the coefficient of the last vector 
involved is bound to be 7. This by itself is sufficient to show 
that we are not dealing with the trivial equation where 
all coefficients are zero. 

We still have to establish the statement just made, 
that the coefficient is 1. 

For definiteness, let us suppose that in the reduction 
of A, B, C, D, • • • , a zero vector appears at the third step. 
We review the process of reduction. Let a, b, c, d, - , 

stand for numbers that appear in the work. These have 
no connection with any symbols previously used . 

We suppose A contains x, perhaps not with coeffi- 
cient 1, but with non-zero coefficient. We can make 
the coefficient of * become 1 by multiplying with a 
suitable constant a. Thus a A — x + • • • . 

We now subtract suitable multiples of this expression 
from B, C, • • • , to get expressions free from x. These ex- 
pressions will be of the form B b(aA), C c(aA), . 

Now we suppose B — b(aA) contains y with non-zero 
coefficient. The coefficient can be made equal to 1 by 
multiplying by a suitable constant d. Thus 

dB — dbaA = y + • • • • 

We now subtract e times the expression above from 
C — caA ; we choose e so that y does not appear in the 
result. We get the expression C — caA — e{dB — dbaA). 

Vectors Over a Field 


Thus it is this expression that we suppose to be zero. 
It will be seen that the coefficient of C is 1. Hence 

C — caA — e(dB — dbaA) = 0 

is a non-trivial equation, which shows that C can be 
expressed in the form fA + gB ; that is, C is a mixture 
of the preceding vectors A, B and does not contribute 
anything new to the space generated. 

This proves our contention for the case where the 
zero vector appears at the third step. There is only 
verbal difficulty in describing the proof for the general 
case. Essentially it is the same as the proof above. 

Question: If a zero vector appears first at the fourth 
stage, show that D is a mixture of A, B, C. 

I shall leave it to you to satisfy yourself that such a 
proof can be constructed for the appearance of a zero 
vector at any stage of the process, and shall regard 
ourselves as having established the following. 

theorem. If a zero vector appears at the m-th stage in 
the reduction of Pi, P 2 , P3, • • • , then P m is a mixture of 
Plj P2, ' " • 5 Pm- 1- 

On pages 145-146 we discussed, from a geometrical 
viewpoint, various ways of finding a basis for a plane. 
Always there seemed to be two vectors in such a basis, and 
it seems plausible that if we choose any two vectors in a 
plane, not in the same direction and neither being zero, 
then these two vectors generate the whole plane. That is, 
it seems that in 2 dimensions any 2 linearly independent 
vectors generate the whole space. In the same way, by 
considering geometrical space of 3 dimensions, it seems 
reasonable that any 3 independent vectors generate the 
whole space. (See page 166 for the geometrical signifi- 
cance of linear dependence.) We are led to conjecture 
the following theorem. 


A Concrete Approach to Abstract Algebra 

theorem. In space of n dimensions , any n linearly inde- 
pendent vectors generate the whole space. 

It may help to see how this is proved if we consider 
a particular example. Linear expressions of the form 
ax + by + cz + dt constitute a space of 4 dimensions. 
Suppose we have 4 expressions of this type, which are 
known to be linearly independent. Then by mixing 
these 4 expressions, it should be possible to obtain any 
expression of the form ax + by + cz + dt. 

Our 4 expressions can be reduced to standard form. 
The expressions are linearly independent; no one of 
them can be a mixture of the preceding ones; hence, by 
the theorem of page 179, it is impossible for a zero vector 
to appear in the reduction. (For if a zero vector did 
appear, it would prove that one of the original vectors 
was a mixture of the others.) Accordingly, the reduction 
must go to its full length. There must be 4 vectors in the 
standard form. 

But there is only one way of getting 4 vectors in the 
standard form, namely by means of the following pat- 

A = x + ey + fz + gt, 

B = y + hz + kt, 

C = z + mt, 

D = t. 

We cannot possibly have a pattern like those on page 172, 
in which long steps appear, and several letters x, y, z, • • • , 
drop out between one line and the next. For if we did 
have this, we would have used up all our letters x, y, z, t 
before we reached the fourth expression, D. Only by the 
utmost economy, dropping one letter only at each line, 
can we reach four lines. (We are compelled to drop one 
letter at each stage. The definition of standard form 
requires it. See the explanation on page 173.) 

Vectors Over a Field 


Thus the standard form A, B, C, D must be of the 
type shown above. (There is, of course, nothing to pre- 
vent some or all of the numbers e, /, g, h, k, m being zero.) 

But now we can clearly obtain any expression ax + 
by -T cz + dt by a suitable mixture of A, B, C, D. By 
taking aA, we obtain the correct coefficient for x. By 
adding a suitable multiple of B to aA, we can make the 
coefficient of y correct, without altering the coefficient 
of x. This would in fact give us a A + {b — ae)B, but the 
details of the calculation are not helpful. By adding a 
suitable multiple of C, we make the coefficient of z cor- 
rect, without disturbing the coefficients of x and y. 
Finally, by adding a suitable multiple of D, we obtain 
the correct coefficient for t. 

This mode of approach generalizes immediately to n 
linear expressions in n variables xi, X2, • • • , x„. The stages 
in the argument are the following. 

(i) The n expressions being independent, the standard 
form must contain n expressions. 

(ii) A standard form with n expressions in it must 
have the form 

A\ = xi + 

A 2 = x 2 + 

A3 — X3 + • • • • 

(iii) Any linear expression a\X\ + a 2 x 2 -f • • • -j- a n x n 
can be gotten by taking suitable values a, c 2 , • • • , c n in 
c\A\ + C2A2 + • • • + c„A„. 

Anyone who has understood the procedure for n = 4 
will see that no new idea is needed to prove the general 

One final objection has to be met. We have taken a 
particular example of a vector space, namely, linear ex- 


A Concrete Approach to Abstract Algebra 

pressions. But the theorem may be needed for some other 
example of a vector space — say, for a geometrical space. 
However, the first few pages of chapter 8 pointed out 
that work with vectors was essentially the same as work 
with linear expressions. 

There are two ways of presenting the proof outlined 
above in a form that shows it to be quite general. 

1. In a vector space of n dimensions, every vector 
is expressible as a\Pi + ciiPi + • • • + a n P n . (See (V.10) 
on page 161, and the explanation of Pi, Pi, • • • , on page 
160.) But this is a linear expression in Pi, Pi, • • • , P n . We 
simply rewrite our earlier proof. Whenever xi occurs we 
write Pi instead. For x 2 we write Pi, and so on. 

2. Alternatively, we may use the label (ai, ai, • • • , a n ) 
for a vector in space of n dimensions, as on page 152. The 
standard form, for 4 dimensions say, will then appear as 

A = (1, e, f, g), 

B = (0, 1, h, k), 

C = (0. 0, 1, m), 

D = (0, 0. 0, 1). 

The argument that any vector (a, b, c. d) can be ex- 
pressed as a mixture of A, B, C, D follows essentially the 
same lines. aA has the first number inside the bracket 
correct. By bringing in B, we can make the second num- 
ber correct also. And so on. The whole proof can be 
carried through in this notation. 

The somewhat informal discussion we have just had 
contains all the ideas needed for the proof of the theorem. 
If you understand these ideas, you can verify that the 
theorem is established. You can, if you like, write out a 
formal proof. I do not give a formal proof in full, since 
it would not be needed by anyone who has understood 
these ideas, and would be meaningless to anyone who 
has not. 

Vectors Over a Field 


We now have our final theorem of this chapter. 
theorem. Any (n + 7) vectors in a vector space of n di- 
mensions are linearly dependent. 

Proof. Let the vectors be V\, V 2 , • • • , F„ , V n +i- It may 
be that Fi, V 2 , • • • , V n are linearly dependent, being con- 
nected by a relation ciVi + C 2 V 2 + • • • + c n V n — 0, 
where ci, • • • , c n are not all zero. If so, we have 

ClVl + C 2 V 2 + • • • + CnV n + 0- V n+1 = 0. 

Since ci, • • • , c n are not all zero, the coefficients in this 
last equation are not all zero; that is, Fi, • • • , V n +\ are 
linearly dependent. 

So the theorem holds if Fi, • • • , F n are linearly de- 

Suppose then that Fi, • • • , V n are linearly independent. 
Then, by the preceding theorem, any vector in the 
space can be expressed as mixture of Fi, • • • , F„. Hence 
F„+i can be expressed as a mixture of Fi, • • • , F„. Hence 
Fi, • • • , V n+ i are linearly dependent. 

Thus the theorem holds in either case, and is proved. 

On page 170 we proved that n linearly independent 
vectors generated a space of n dimensions. Why then, it 
might be asked, did we find it necessary to prove on 
pages 180-182 that, in a space of n dimensions, any n 
linearly independent vectors generated the whole space? 
Is it not obvious that they generate the whole space? 
For instance, someone might say, “take the case n — 2. 
We have a plane; we take any two independent vectors; 
they generate a plane — isn’t it obvious that this plane 
must be the plane we started with?” It is indeed plausible 
that it should be so. However, the chain of reasoning 
just given assumes a certain result, namely that one 
plane cannot be contained in another plane — this to be 
understood in the sense that you cannot remove certain 
points from a plane, and still have a complete plane re- 


A Concrete Approach to Abstract Algebra 

maining. More generally, if S and T are spaces of n 
dimensions, and T is contained in S, then T is identical 
with S. If one w-dimensional space lies in a second 
n-dimensional space, then it completely fills it. 

Now this result sounds very reasonable, but we have 
to prove it. And, in effect, that is what we did on pages 

On page 148 the question was raised whether a space 
might be at one and the same time a space of n dimen- 
sions and a space of (n + 1) dimensions. It seemed un- 
likely, but again unlikely things do happen — for instance, 
there is a curve that completely fills a square; it is as 
well to have a proof. The last theorem of this chapter 
supplies the proof. In space of (n + 1) dimensions there 
are n + 1 independent vectors — for example the vectors 

Vi = (1, 0, 0, 

• • • , 0), 

Vz = (0, 1, 0, 

• • , 0), 

Fa = (0, 0, 1, 

• • , 0), 

n+l = (0> 0> 

In space of n dimensions there cannot exist (n -f- 1) inde- 
pendent vectors. Hence it is impossible for a space of 
n dimensions to be a space of (n + 1) dimensions. 

Chapter io 

Fields Regarded as Vector Spaces 

We have now proved some results for vector spaces, 
but it is far from evident that these results will tell us 
anything interesting about fields. In fact, some very use- 
ful theorems about fields can be obtained. 

To see how this happens, we review the procedure of 
modern mathematics. The underlying idea of modern 
mathematics is to extract the essence of any proof, and 
separate it from purely accidental aspects. In this way 
generality is obtained. 

If you want to write a paper in modern algebra, you 
look at a theorem somebody else has proved, and you 
see just what he has used to prove it. For example, the 
theorem that any quadratic has at most two roots was 
originally proved for numbers. We examine the proof 
and ask, “What properties of numbers does it use?” We 
find that it uses only the properties of numbers expressed 
by axioms (7) through (72). This means that the proof 
can be written in such a way that it applies to any set of 
elements — not necessarily numbers — that obey axioms 
(7) through (72). 

So, without having used any real originality, we can 
state a more general theorem: “In any structure, the 


A Concrete Approach to Abstract Algebra 

elements of which obey axioms (7) through (72), a quad- 
ratic has at most two roots.” 

We can now shorten this statement by bringing in a 
technical term, “field.” Any structure, we say, for which 
axioms (7) through (72) hold, will be referred to as a 
field. Our theorem then takes the form, “In any field a 
quadratic has at most two roots.” 

A most important point should be noticed here; the 
misunderstanding of it frequently causes confusion. The 
qualification for being a field is positive, not negative. 
That is to say, to be a field, a structure must satisfy 
axioms (7) through (72). It is not disqualified from being a 
field because it obeys other axioms in addition to (7) through (72). 

There is an obvious reason for taking it this way. The 
proof that a quadratic equation has at most two roots 
uses axioms (7) through (72) only: so it applies to any 
structure for which these axioms hold. The proof no- 
where requires that the structure does not satisfy other 
axioms. Real numbers, for instance, do satisfy other 
axioms. There is the relation a < b, “ a less than b,” 
which is defined, and for which a system of axioms holds; 
for example, it is an axiom that if a < b and b < c, then 
a < c. Not all fields permit such a relation to be defined; 
you would, for instance, find difficulty in defining an 
order relation for the arithmetic modulo 5; again, we do 
not use the symbol < for complex numbers. However, 
the symbol < does not occur anywhere in the proof that 
a quadratic has at most two roots. For the proof of this 
theorem, it is entirely irrelevant whether the structure 
has a relation a < b or not. 

Thus, it is correct to say, “The axioms for a field do 
not require a < b to be defined.” It would be incorrect 
to say, “No structure in which a < b is defined is a field.” 
In the same way the definition of “human” does not 
require the possession of unusual physical strength, 

Fields Regarded as Vector Spaces 


American citizenship, or the wearing of a hat. But a 
strong, American, hat-wearing individual is still to be 
regarded as human; and all theorems that can be de- 
duced from the axioms specifying humanity will be true 
for such an individual. We do not think of “a human 
being,” “a strong creature,” “an American,” “a hat 
wearer” as being distinct objects. Rather these terms 
emphasize different aspects of objects — or, it may well 
be, different aspects of one single individual. 

Thus, there is no contradiction between our having 
proved, in exercise 7 on page 31, that the numbers 
a + by/ 2 (with a, b rational) form a field, and on page 1 54 
that they form a vector space of 2 dimensions over the 
rationals. The usual objection is, “You can multiply 
a + by / 2 and c + dV 2, but you cannot multiply two 

The mistake here is again that of using the tests nega- 
tively rather than positively. The basic property of a 
vector space of 2 dimensions is that every element of it is 
a mixture of two basic ingredients only. Now every num- 
ber a -f- by/ 2 is a mixture of 1 and y/l. By using this 
property (in the form of the more precise statements we 
had earlier), we can prove for the numbers a + by/ 2 all 
the theorems appropriate to spaces of 2 dimensions. 

The fact that y / 2 can be multiplied by V2 corresponds 
to the aspect of this structure as a field. We cannot deal 
with this aspect by means of the vector axioms. 

The useful fact is that a + by/ 2 is simultaneously a 
vector space and a field. We can prove for it both the 
theorems appropriate to vector spaces and to fields. In 
this way we may be led to new results, obtained by com- 
bining these theorems. 

In terms of our earlier illustration, we suppose it a 
theorem that human beings are conscious of what is 


A Concrete Approach to Abstract Algebra 

happening, and that a hat- wearer’s head is warm. We 
can deduce that a human being wearing a hat is con- 
scious of warmth in the head. This result unites the two 
theories: it cannot be proved in either theory alone. A 
human being may have a cold head in the absence of a 
hat. A lower animal, although wearing a hat and hence 
having a warm head, may not be conscious of the 

There is a property of the numbers a + b V2 that will 
serve to illustrate this theme, and lead us to a more gen- 
eral result. 

Every number of the form a -f- bV 2 satisfies an equa- 
tion, with rational coefficients, which is quadratic (or 
simpler in particular cases). For if 

x = a + bV 2, 
x — a = 2, 

(x — a) 2 = 2b 2 . 

Hence x 2 — 2 ax + a 2 — 2 b 2 = 0. 

This is an equation with rational coefficients. 

This method works all right in this particular case, but 
it does not generalize easily. For instance, it is true that 
every number a + b^/ 2 + c's/A satisfies a cubic equa- 
tion over the rationals, and that every number a + bV 2 
+ cV 3 + 6 satisfies an equation of the fourth degree 

over the rationals; a, b , c, d of course represent rational 
numbers. One can easily construct more and more com- 
plicated examples of such theorems. The simple method 
used above for a + iV 2 does not enable us to see the 
truth of these more complicated theorems. We accord- 
ingly look for a method that will generalize. 

We return, then, to proving that a -\- bV 2 satisfies a 
quadratic and look for a more illuminating proof. 

A quadratic expression px 2 -fi qx + r may be written 

Fields Regarded as Vector Spaces 


P' x 2 + <l’ x + that is to say, it is a mixture of x 2 , x , 
and 1. We want to show that, for some rational numbers 
p, q, r, this mixture will be zero. Let us take a particular 
case, say x = 3 + 5V2. What are x 2 , x, and 1? 

1 =1, 

x = 3 + 5\/2, 
x 2 = 59 + 30V2. 

We seek to make a mixture of these three things zero. 
Now, from the equations above, 

px 2 + qx + r = (59 P + 3 q + r) + (30 p + 5q)Vl. 

As 59 p + 3q + r and 30 p + 5 q are rational if />, q, r are 
rational, we can only make the above mixture of 1 and 
V2 zero by making 

59 p + "bq + r = 0 (1) 

30 p + 5? =0 (2) 

Here we have two equations in the three unknowns 
/>, q , r. Our theorem will be proved if we can show (i) 
that non-trivial solutions exist, (ii) that some such solu- 
tion makes p , q, r rational. The trivial solution p = q — 
r = 0 of course is useless. 

In this particular case, we can solve the equations, 
and verify that the solution is rational. However, we are 
seeking to extract some general principle. 

The language of vector spaces helps us to do this. 
px 2 + qx + r • 1 =0 with rational p, q, r is another way 
of saying that * 2 , *, 1 are linearly dependent over K, 
the rationals. 

Now, if 1, V2 are chosen as a basis, x 2 has the label 
(59, 30); x has the label (3, 5); 1 the label (1,0). 

Equations (1) and (2) can be combined in the vector 

PC 59, 30) + q( 3, 5) + r(l, 0) = (0, 0). 


A Concrete Approach to Abstract Algebra 

But we know that any 3 vectors in 2 dimensions are 
linearly dependent. So we can be sure that numbers 
p, q, r, not all zero, exist which satisfy the above equation. 

We need these p , q, r to be rational. Gan we be sure of 
fulfilling this condition? We are working in the frame- 
work of Example 4 on page 154. K is the field of rational 
numbers. The numbers a, b that appear in any label 
(a, b) are rational. The linear dependence of vectors 
A, B, C means that pA + qB + rC = 0 with p, q , r ele- 
ments of K, not all zero. All the work proceeds within the 
field K. It may help you to see this, if we work through 
the proof that (n + 1) vectors in n dimensions are linearly 
dependent, applying its procedure to the particular case 
we have here, namely that the three vectors 
A = (1, 0), 

B = (3, 5), 

C = (59, 30), 

in 2 dimensions are linearly dependent. We should first 
express A and B in standard form; this gives us two vec- 
tors which form a basis for the whole space. Hence C is 
expressible as a mixture of these. In fact, 

B - 3A = (0, 5). 

D = iB - iA = ( 0 , 1 ). 

A and D make the standard form. Evidently 
C = 59A + 30 D, 

= 59 A + 30 (iB - f A), 

= 41 A + 6 B. 

The numbers 1, 0, 3, 5, 59, 30 that occur in the labels 
for A, B, C are necessarily rational. In the calculation, 
we start with these numbers and perform addition, sub- 
traction, multiplication, and division, but no other operation 
is used. Accordingly, we stay always within K, the field 
of rationals. We are bound to arrive at a rational solu- 
tion p, q, r. 

Fields Regarded as Vector Spaces 


In this particular case, C — 6B — 41 A = 0 : so p — 1, 
q = — 6, r — —41, and x = 3 + 5\/2 satisfies x 2 — 
6x — 41 =0. This, of course, is the same equation that 
we should have obtained by applying our first method. 
For this particular case, the second method is longer. Its 
value lies in the fact that it applies to other cases. The 
purpose of the method, too, is not so much to calculate 
the equation, as to show that an equation exists. 

The work above shows that the theorem “Any (n + 1) 
vectors in n dimensions are linearly dependent” can be 
stated in elementary algebra. For suppose the vectors are 

Vi with label (a h b h a, • • • , gi), 

V 2 with label (a 2 , b 2 , c 2 , • • • , g 2 ), 

Fn-j-i with label (a n+i, rn-j-i, 3 ^71+1)3 

where all the numbers a\, • • • , g n+ i belong to a field K. 
The linear dependence of these (w + 1) vectors means 
that there are numbers pi, p 2 , • • • , p n + 1, belonging to K 
and not all zero such that p\V\ + p 2 V 2 + p n+ \ V n+ i = 0. 
That is, 

O-lpl + a 2 p 2 -T * * * + On+lpn+i = 0, 

b\p\ + b 2 p 2 + • • • + bn+ip n +i = 0, 

g\p\ + g2p2 + ’ ' * + gn+lpn+1 = 0 , 

so we have the theorem. If n equations in (n + 1) unknowns, 
of the type shown above, have all their coefficients lying in a field 
K, then these equations have a solution pi, p 2 , • • • , p n+ i com- 
posed of elements of the field K, not all zero. 

It is essential to include the words “of the type shown 
above.” For instance, the three equations in four 

x+jy+ z + r = 1, 
x + 2y + 3z + 4r = 2, 

3x + ly + 12* + 17r = 3, 


A Concrete Approach to Abstract Algebra 

have no solution, as is easily verified if you regard r as 
known, and try to solve for x, y, z. But these equations 
are not of the type specified, since they have non-zero 
constant terms 1, 2, 3. 

This example also shows that the theorem is not trivial. 
Having fewer equations than unknowns does not always 
guarantee the existence of a solution. 

We now use the method developed above to show that 
x = a + b\/ 2 + 4, (a, b, c rational), satisfies an equa- 

tion px % + qx 2 + rx + s = 0, where p, q, r, s are not 
all zero. 

We noted on page 121 that the numbers t + u\/ 2 + 
v\^4, with t, u, v rational, form a field. Since x belongs 
to this field, x 2 and x 3 also belong to the field, by field 
axiom (2). We could work out x 2 and x 3 in terms of a , b, c, 
but this work would be wasted. All that matters for the 
latter part of the proof is that x 2 and x 3 are of the form 
t + mV 7 2 + flV 7 4. Accordingly, we write 

1 = !, 

x = a + b^/ 2 + r'V / 4, 
x 2 = d + eV2 + fy/ 4 , 
x 3 = g + h\/ 2 + ky/ 4, 

where d , e, /, g, h, k could be expressed in terms of a, b, c, 
but all we really need to know is that these symbols 
represent rational numbers. 

We could now consider px z + qx 2 + rx + s and show 
that it would be zero if the four quantities p, q, r, s sat- 
isfied the three equations 

pg + qd + ra + s = 0 , 
ph qe + rb = 0 , 

pk + qf + rc — 0 . 

Fields Regarded as Vector Spaces 


We know that three such equations in four unknowns 
always have a rational solution. Hence we know that 
there are rational numbers p , q, r, s, not all zero, for 
which px 3 + qx 2 + rx + s = 0, and our theorem is 

In order to explain this proof, I have had to bring in 
symbols d, e, f, g, h, k. But really I have not made any 
calculations with these symbols. If I wanted to prove that 
a + b's/ 2 + c\/ A + d \/ 8 + eX^ 16 (a, b, c, d, e rational) 
always satisfies an equation of the fifth degree over the 
rationals, I should pretty well use up the whole alphabet 
in explaining the proof. But the proof would not contain 
any new idea. It is therefore natural to coin a name for 
such a situation, to have a brief way of indicating that 
this proof works. That name we already have: “vector 

Our proof that every number a + b's/ 2 -f- cs/ A satis- 
fies a cubic equation can now be stated very briefly. 
The rational numbers being denoted by K, the totality 
of numbers t + us/2 + vX^A, with t, u , v rational, con- 
stitutes the field K(\ // 2). 

We now argue let x = a + b's/ 2 + c's/ 4. x is an ele- 
ment of K(y / '2 ) . Since 2) is a field, x 2 and * 3 are also 

elements of K{y/ 2). 1 is an element of K^T). 

Hence 1, x, x 2 , x 3 are elements of iC(V / 2). 

But every element of K(s/2) is a mixture of 1, '\/2, 
</ 4, with rational coefficients. Hence K{' s/ 2) is a vector 
space of 3 dimensions* over K. Hence any four elements 
of it are linearly dependent over K. Hence, for rational 

* 1, 2 , vT form a basis since they are linearly independent 

over K. But we do not even need to prove this. If they were 
linearly dependent, K{ ^2) would be a space of less than 3 dimen- 
sions, and the proof would be even stronger than it is now. 


A Concrete Approach to Abstract Algebra 

numbers p, q , r, s, not all zero, we have px z + qx 2 + 
rx + s = 0. Q.E.D. 

Question 1: If w stands for the real number such that 

w 7 = 2 and s = a + bw + cw 2 + dw 3 + ew 4 + f wh + 
where a, b , c, d , e> /, g are rational numbers, prove that 
s satisfies an equation of the seventh degree with rational 
coefficients. (This question is intended to show the econ- 
omy of thought and statement made possible by using 
the vector terminology, as compared with the notation 
of elementary algebra.) 

Question 2: Do the elements a + bx / 2 + cV 3 + dx / 6 

form a vector space over the rational numbers K? (a, b, 
c, d rational numbers) . Of what dimension? What is the 
degree of the equation with rational coefficients satisfied 
by the general element of this correction? 

Question 3: Let m — V2 + V^. Calculate m 2 and m 3 . 

Do 1, m, m 2 , m 3 form a basis for the space considered in 
question 2? What is the simplest equation over the ra- 
tional that is satisfied by m ? What are the other roots of 
this equation? (The equation for m can be found quite 
easily by elementary algebra.) 

Question 4: An irrational number r satisfies an equa- 

tion f(x) = 0, which is of degree n and irreducible over 
the rational numbers K. Find a basis for K(r), regarded 
as a vector space over K. Of how many dimensions is it? 
What is the degree of the equation with rational coeffi- 
cients satisfied by an arbitrary element of K(r)? 

In our examples, we have taken K as the field of ra- 
tional. Often indeed, we are interested in knowing 
whether an element satisfies an equation with rational 
coefficients. But we might also be interested to know 
whether it satisfied an equation with roots drawn from 
some other field, F. 

Fields Regarded as Vector Spaces 


We have the following general theorem. Suppose F is any 
field , and G is a field that contains F. Suppose G is a vector space 
of n dimensions over F; that is to say, there are fixed ele- 
ments bi, b 2 , • • • , b n of G such that every element of G 
can be represented in the form afb\ + a 2 b 2 + • • • + a n b n 
by choosing suitable elements a h a 2 , • • • , a n of F. Then 
every element of G satisfies an equation of degree n {or less) 
over F. 

For instance, if R denotes the field of the rationals, F 
might be Z?(V: 2) and G might be F(V3). That is, we 
start with the rationals, R. F is obtained by bringing in 
the new element V2; anything that can be gotten by 
adding, subtracting, multiplying, and dividing with ra- 
tional numbers and Vz belongs to F. In fact, every ele- 
ment of F can be expressed in the form p + q's/ 2, with 
rational p , q. We use the letters u, v to denote elements 
of F. 

G is obtained from F by allowing the free use of the 
extra symbol v^3. In fact, every element of G can be put 
in the form u -f- &V 3, where u, v belong to F. 

Thus, in this example, b\ = 1 and b 2 = V3 are the 
fixed elements of G. a\ = u and a 2 = v are variables, 
representing elements of F. Thus G is a vector space of 2 
dimensions over F, and every element of G satisfies an 
equation of the second degree — that is, a quadratic — 
with coefficients in F. 

For example, x = 'V / 2 + V / 3isan element of G, with 
a\ — V2 and a 2 = 1. 

Evidently (x — V2) 2 = 3, which simplifies to 
*2 _ 2 V2 x - 1 = 0. 

This is a quadratic equation over F. The coefficients 
are 1, — 2V2, — 1, all of which lie in F. It is not a 


A Concrete Approach to Abstract Algebra 

quadratic over the rationals R, since — 2V2 is not in R. 

If we wish to obtain the equation over R satisfied by 
Vl + V3, we write the quadratic above in the form 

x 2 - 1 = 2V2 x. 

Squaring gives 

(x 2 - l) 2 = 8x 2 , 


x 4 - 10x 2 + 1=0. 

This equation is of the fourth degree, which is quite 
consistent with our theorem. For in u + v V 3, the general 
element of G, the quantities u and v are elements of F; 
that is to say, we may write 

u = c + dy/ 2, 
v — h + ky/ 2, 

where c, d, h, k are rational. 

Thus, the general element of G is 

(c + dV 2) + (h + kV 2) V3 

c + dV 2 + hV 3 + kV 6. 

G is thus a vector space of 4 dimensions over R, with 
1, V2, V3, y/ 6 as a basis. Accordingly, any element of 
G must satisfy an equation of the fourth degree at most, 
with rational coefficients. Thus, 

G is a vector space of 2 dimensions over F. 

G is a vector space of 4 dimensions over R. 

Any element of G satisfies a quadratic equation if 
coefficients from F are allowed, but an equation of the 
fourth degree is all we can guarantee if only rational 
coefficients are allowed. 

Fields Regarded as Vector Spaces 


Finding the Equation of Lowest Degree 

for Any Element 

We have already met the principle that x satisfying an 
equation of degree n over any field K is the same as 
1, x, x 2 , • • • , x n being linearly dependent over K. (See 
pages 189-194.) 

Thus, if x satisfies an equation of degree n, but no 
equation of degree less than n, then 1, x, x 2 , • • • , x" will 
be linearly dependent, but 1, x, x 2 , • • • , x n_1 will be 
linearly independent. 

Thus, it is possible to find the value of n by forming the 
quantities 1, x, x 2 , x 3 , • • • , and seeing at what point linear 
dependence first happens. The standard form for vectors 
gives a systematic way of doing this. The appearance of 
a zero vector is the signal for linear dependence having 

Example. Find the degree of the simplest equation 
over the rationals satisfied byx=v / 2 + 'V / 3. 

It saves rearrangements if we take our basis in the 
order 1, Js/2, V6, V3 so that (a, b, c, d ) is the label for 

a + bV 2 + C V 6 + dV: 3. 

1 has label (1, 0, 0, 0), 

x = V2 + v/ 3 has label (0, 1, 0, 1), 

x 2 = 5 + 2V"6 has label (5, 0, 2, 0), 

x 3 = 11 V2 + 9V3 has label (0, 11, 0, 9), 

x 4 = 49 + 20V6 has label (49, 0, 20, 0). 

There is no point in calculating further. We now have 
five vectors in 4 dimensions. These must be linearly 
dependent; but, conceivably, linear dependence might 
have arisen even before x 4 was reached. 

The vectors of the standard form are 


A Concrete Approach to Abstract Algebra 

1 with label (1, 0, 0, 0), 

x with label (0, 1, 0, 1), 

|(x 2 — 5) with label (0, 0, 1, 0), 

— ^(x 3 — llx) with label (0, 0, 0, 1). 

No zero vector has appeared, so 1, x, x 2 , x 3 are inde- 
pendent and generate the whole space, x 4 is 49 times the 
first vector added to 20 times the third vector of the 
standard form above. That is, 

x 4 = 49-1 + 20-K* 2 “ 5), 

whence x 4 — 10x 2 + 1 =0 as we expected from our 
earlier work. We have now demonstrated, by using a 
vector procedure, that x does not satisfy any simpler 
equation over the rationals. 

This type of example is included here not so much be- 
cause we wish to perform calculations of this kind, but 
so that you can become used to thinking of elements of 
fields as vectors. This idea is important for the general 

Question: Find the degree of the simplest equation 

over the rationals satisfied by V / 3 — V" 2. Also find the 

Must a Field Have a Dimension 

Over a Sub-field ? 

Often one field G contains another field F. (Then F 
is called a sub-field of G. Thus the rationals are a sub- 
field of the reals because the reals contain the rationals.) 
We have had examples in which G was a vector space of 
n dimensions over F. Must this always happen? If a field 
G contains a sub-field F, must G be a vector space of n 
dimensions over F ? 

We must first get one difficulty out of the way. Sup- 

Fields Regarded as Vector Spaces 


pose, for example, that G is the real numbers and F the 
rationals. Now the numbers V2, V3, V5, V7, • • • , the 
square roots of the prime numbers, are all real numbers, 
hence in G. But they are linearly independent over the 
rationals (this can be proved; you should find it reason- 
able that, for example, Vll cannot be expressed as a 
rational mixture of V2, V 3 , "n/ 5, and V7). Thus G 
contains as many elements as you like that are linearly 
independent over F. Now a space of n dimensions cannot 
contain (n + 1) independent elements. So G cannot have 
any finite dimension over F, in this case. 

Accordingly, in any theorem we obtain, we must in- 
clude some condition to rule out the possibility that G 
contains infinitely many elements independent over F. 

Is there anything left to investigate? Will not our 
theorem have to run, “If G is of finite dimensions over 
F, then G is of n dimensions over F for some integer n,” 
which is futile? No, there is still a possibility to consider. 
For instance, is it possible for G to be something more 
than a space of 2 dimensions over F, but less than a space 
of 3 dimensions? 

What would this mean? Let small letters, a 0 , a\, « 2 , * • • , 
denote elements of F, while capital letters Pi, P 2 , • • • , de- 
note elements of G. “G is something more than a space 
of two dimensions over P.” This means that, if we take 
the two ingredients 1, Pi, all possible mixtures + 
ai'Pi give elements of G, but these do not completely 
cover G. There are some elements of G that cannot be 
represented in this way. It is of course assumed that 1, Pi 
are linearly independent over F. On the other hand, G 
is something “less than a space of 3 dimensions over F.” 
This means that taking three elements 1, Pi, P 2 would 
give us too much. The elements ao • 1 + a\ • Pi + • P 2 

give all the elements of G, and some others besides . 


A Concrete Approach to Abstract Algebra 

The question is, is such a state of affairs possible? In 
fact it is not possible. We can prove that if G is more 
than a space of 2 dimensions over F, then it is at least a 
space of 3 dimensions over F. 

We have 1, P x linearly independent over F, and 
0 o • 1 + fir Pi always gives an element of G. But some 
element of G cannot be expressed in this form. Choose 
any such element; call it Pi. 

First result. 1, Pi, P 2 are linearly independent over F. 
For, if not, there are elements of F, c 0 , c%, c 2 , not all zero, 
such that co'l + c X 'P\ + r 2 *P 2 = 0. If ci 0, we can 
divide by c 2 and solve for P 2 . This gives P 2 as a mixture 
of 1 and Pi. But we supposed P 2 outside the elements of 
the form co'l + a.yP\. So c 2 = 0. But this means that 
1, Pi are linearly dependent with cq-\ + ci-Pi = 0. And 
this also is ruled out. Accordingly, it is impossible that 
1 5 Ph P 2 should be linearly dependent over F ; our result 
is proved. 

This shows that 1, Pi, P 2 generate a space of 3 dimen- 
sions. They do not lie in any space of 2 dimensions (for 
no 2 dimensional space can contain 3 independent 
vectors) . 

Thus the elements a 0 - 1 + arPi + a 2 -P 2 do form a 
space of 3 dimensions. We proceed to show that this space 
lies completely inside G. 

Second result. Every element of the form ao • 1 + fli*Pi 
+ < 22 ‘Pi belongs to G. 

Proof . By our assumptions, a 0 • 1 + a x • Pi is an element 
of G for all a 0 , a x of F. P 2 is an element of G (see how it 
was defined just before the statement of “first result”), 
ai is an element of F; since F is contained in G, then a 2 
belongs to G. Thus, ao-\ + a x • P x , a 2 , P 2 are three ele- 
ments of G. Addition and multiplication of elements of 
a field always give elements of that field. But a Q -\ + 
at Pi + di-Pi is obtained from the three elements of G 

Fields Regarded as Vector Spaces 


just listed by multiplication and addition only. Hence, 
this expression is an element of G, as had to be proved. 

Thus G certainly contains a space of 3 dimensions 
over F. 

The above proof does not use any properties peculiar 
to the numbers 2 and 3. We can show, without using any 
new idea, that a field never lies between one whole num- 
ber and the next in its dimension over a sub-field. 

Question: The field G contains the field F. G contains 

a vector space of n dimensions over F, but these are ele- 
ments of G not belonging to this vector space. Prove that 
G contains a vector space of (n + 1) dimensions over F. 

It is now easy to see how things work out. A field G 
contains a sub-field F, that is, every element of F is in G. 

It may be that G coincides with F. If so, G is a space 
of 1 dimension over F, with 1 as a basis. 

But there may be an element Pi in G but not in F. 
Then G certainly contains the vector space of 2 dimen- 
sions over F, with basis 1, P x . This space may fill G. Then 
G has 2 dimensions over F. 

If it does not fill F, there is an element P 2 of G outside 
it. G then contains the 3-dimensional space with basis 
lj Pi) P 2 . If this space fills G, then G is of 3 dimensions 
over F. If not, we bring in P 3 . 

So we continue. How can the process end? Well, first 
of all, it may not end. We may be able to go on forever 
bringing in new elements. 

This is case (i) ; for every n, we can find 1 , Pi, P 2 , • • • , 
P n -h n elements of G linearly independent over F. 

On the other hand, the process may come to an end. 
We obtain, for some n, a space contained in G with basis 
1) -Pi) P 2 ) • • • , Pn-i, but we are unable to make the next 
step. What can be preventing us? All that we need for 
the next step is some element not in the space generated 


A Concrete Approach to Abstract Algebra 

by the basis 1, Pi, P 2 , • • • , P„_i. Our inability to proceed 
can only mean that there is no such element. That is, the 
field G coincides with the space; thus G is a space of n 
dimensions over F. This is case (ii). We thus have the 
following theorem. 

theorem. If F is a sub-field of G, either (i) there is no limit 
to the number of elements of G linearly independent over F, or (ii) 
G is exactly a vector space of n dimensions over F, where n is 
some natural number. 


1. If R is the rationals, F = R(\/ 6), G = P(V / 2, V^), show 
that F is a sub-field of G, and find the dimension of G over F. 

2. It is known that it does not satisfy any equation with 
rational coefficients. If F = R, the rationals, and G = R(tt), 
does case (i) or case (ii) of the above theorem apply? 

3. Find the dimension of R(x / '2) over P(V / 2), R being the 

4. Find the dimension of R (\/ 1 + V5) over R(V 5). 

5. Find the dimension of R ( 1 + V2) over R(s/ 2). 

6. Find the dimension of GF( 2 2 ) over GF( 2), and of GF( 2 s ) 
over GF( 2). (See page 130 for terminology.) 

Repeated Extensions of a Field 

We have already had examples in which a field is 
repeatedly extended. For example, by adjoining Vz to 
the rationals R, we obtain S = P(V 2). By adjoining V 7 5 
to S, we obtain T = S('\/ 5) = P(v / 2, 'S/S). T contains 
1, 3^5, V / 25. It also contains 1, V2. It must contain the 
products of these elements, which we can arrange nat- 
urally in a rectangle. 

Fields Regarded as Vector Spaces 


1 V2 

\/ 25 V2-v / 25 

These six quantities are linearly independent over R, 
and any element of T is a mixture of these with rational 
coefficients. Thus T is of 6 dimensions over R. S is of 2 
dimensions over R. 

The arrangement of the 6 quantities in a rectangle 
emphasizes that 6 is 3 times 2. It suggests that such an 
arrangement would be possible in any extension of S; 
that the basic quantities could always be written 

1 V2 

A AV2 

B BV 2 

M MV 2 

There would thus be an even number of elements in 
the basis. This suggests the theorem: “Any extension of 
R(y/ 2) has dimension 2 n over R, for some integer 
It is easy to verify that you cannot have an extension 
oiR(V2) of 3 dimensions over R. For suppose you could. 
Let 1, V2, Q be a basis of such a field. 1, V2, Q must be 
linearly independent over R; otherwise they will not form 
a basis for a space of 3 dimensions. The product QV 2 
must be in this field. Hence, it must be a mixture of 
1, V2, Q. That is, for some rational a, b, c, 

QV 2 = a + bV 2 + cQ. 


Q(V 2 - c) = a + bV 2. 

Now V2 — c is not zero, for c is rational while V2 is 
irrational. Hence, we can divide by V2 — c Both 


A Concrete Approach to Abstract Algebra 

^2 — c and a + b^/ 2 lie in the field i?(V2). Hence, 
their quotient also lies in this field. Thus, for some ra- 
tional d , e 

Q = d+ eV2. 

But this shows Q to be a mixture of 1, V2, which contra- 
dicts our assumption that 1, V2, Q are linearly inde- 

The position is this. You cannot get a field by intro- 
ducing one new element Q into the basis. Either Q is a 
mixture of 1 , V2, which adds nothing new, and you still 
have a space of 2 dimensions over R; or Q and Q V2 both 
make a contribution and you have a space of 4 dimen- 
sions with the basis 

1 V2 

Q QV 2 

You can of course have a field of 3 dimensions over R . 
1, V^, V4 is the basis of such a field. But this field, R{\^ 2), 
does not contain V2. So it is not an extension of R(\ / '2). 

There is nothing special about square roots in this 
connection. R(\ // 2) has dimension 3 over R. Any exten- 
sion of R{'\/2) will have dimension 3 n over R. For in- 
stance, R(V2, has the basis 



</ 4 

\ // 3 ■ 



</ 9 ■ 


\ // 9 ■ 



21 • 

• V / 2 

</21 • 



^ 81 - 


and forms a space of 1 5 dimensions over R. 1 5 is a mul- 
tiple of 3. 

We may now state the theorem of which the above 

Fields Regarded as Vector Spaces 


have been examples. It is convenient to use the abbrevi- 
ation P C Q for “P is contained in Q,” that is, every 
element of P is an element of Q. (P may coincide with Q, 
but usually will be a part, not the whole, of Q.) 
the ore m . Let F, G , H be fields, F d G C H. Let G be a 
vector space of p dimensions over F, and H a vector space of q 
dimensions over G. Then H is a vector space of p-q dimensions 
over F. 

In the first example of this section, F = R, G — S, 
H = T, p = 2, q = 3, and T is of 2 • 3 = 6 dimensions 
over R. 

The proof of this theorem is extremely simple. It in- 
volves no calculations. It requires only an understanding 
of such ideas as linear dependence over a particular field. 
It may help you to follow the proof if you apply it to the 
particular examples we have had. 

Proof. It will be convenient to use small letters, 
a, b, c, • • • , for elements of F, and capitals A, B, C, • • • , 
for elements of G. Capitals from the end of the alphabet 
will be used for elements of H, say U, V, W, . 

We now start to express concretely the information 
at our disposal. 

“H is a vector space of q dimensions over G.” This 
means that every element of H is a mixture of q basic 
ingredients. The coefficients, which show how much of 
each ingredient is used, are drawn from G. But the in- 
gredients themselves belong to H. (Look back to earlier 
examples. So far as the present statement is concerned, 
it is irrelevant that G contains a sub-field F. One could 
take as an example R for G and P( V 2) for H. This ex- 
ample of course will not do for the later part of the work.) 

Thus every element of H can be expressed in one and 
only one way as 

V = AiUi + AfiU, + • * • + A q U q 

( 1 ) 

206 A Concrete Approach to Abstract Algebra 

where £/i, • • • , U q are fixed elements of H, while 
Ai, • • • , Ag are variables over G. The elements 
Ui, • • • , Uq , since they generate a space of q dimensions, 
are linearly independent over G. V = 0 only if A\ — 
Ai = • • • = Aq — 0. 

Now G in its turn is a vector space of p dimensions 
over F. Every element of G can be expressed as 

B = ciD\ 4~ C 2 D 2 4" * * * H - CpDp (2) 

where Di, • • • , D v are fixed elements of G, linearly in- 
dependent over F, and c h • • • , c v are variables over F. 

Now A 1 , • • • , Aq are elements of G, and hence each of 
them can be expressed as a mixture of D 1 , • • • , D p with 
coefficients from F. Suppose, then, that 

A\ = cnDi + C 12 D 2 + • • 

• 4- ClpDp 

A 2 — C 21 D 1 + C 22 D 2 + * • 

• 4" c2pD p 


Aq = C q lDl 4" C q %D2 4“ 1 * 

4“ C qpDp. 

If we substitute these values in the expression for 

V and 

multiply out, we find that 

V — cn U\D\ 4" cuU\D2 4~ * 

• • 4“ c\ v U\Dp 

4~ C 21 U iD\ 4- C 22 U 2 D 2 4- * 

• • 4“ C2pU2Dp 



4“ C q \UqD\ 4“ Cq2U q D2 4~ * 

• • "1" CqpXJqDp* 

We now have V expressed as a mixture of the pq fixed 
quantities U\D\, • • • , U q D p . The coefficients cn, • • • , c qp 
are elements of F. V is any element of H. Thus H cer- 
tainly cannot be a vector space of more than pq dimen- 
sions. It will be a vector space of less than pq dimensions 
if UiD h ’ • • , UqD p are linearly dependent over F. If we 
can show these to be linearly independent, we shall 
know that they generate a space of exactly pq dimensions. 

If they were linearly dependent, it would be possible 

Fields Regarded as Vector Spaces 


to find city • • • , c qp , not all zero, in such a way that V, in 
equation (4) would be zero. 

But we have already seen, just below equation (1), 
that V = 0 only if A\ = Ai — • • • = A q — 0. 

Further, JDj, • • • , D v are linearly independent over F. 
In equations (3), A\ = 0 only if cn = c\% = * • • = 
ci P = 0; A 2 = 0 only if C 21 = C 22 = • • • = C 2 P = 0, and so 
on through A q = 0 only if c q \ = c<$ = • • • = = 0. 

Thus F = 0 only when all of cn, • • • , are zero. This 
means that U\D\, • • • , Z7 g -D p are linearly independent. 
Accordingly, they generate a space of pq dimensions. 

We have shown that every element of H lies in this 

We must also show that every element of the space lies 
in H. But all the quantities cn, • • • , c qp ; Ui, • • • , U q ; 
Di, • • • , D p belong to H. In equation (4) they are com- 
bined by addition and multiplication only. Since H is a 
field, the result must lie in H. Thus, whatever values are 
chosen for cn, • • • , c qp , the right-hand side of equation (4) 
gives an element of H. 

Thus every element V has a label (cn, • • • , c qp ), and 
every such label belongs to some element V of H. 

H is thus a vector space of pq dimensions over F. The 
theorem is proved. 

Chapter a 

Trisection of an Angle 

Before going into details we examine, in broad outline, 
the proof that angles cannot be trisected. The meaning 
of the statement should first be clarified. There are plenty 
of mechanical devices for trisecting angles. Trisection is 
impossible only within the rules proposed by the ancient 
Greeks — compass and straightedge alone permitted as 
implements, and these to be used only in the ways cus- 
tomary in Euclidean geometry. 

Certain angles, of course, can be trisected. The angles 
30° and 15° can be constructed; so it is certainly possible 
to trisect 90° and 45°. But, in general, given an angle, 
there is no procedure for trisecting it. 

If there were a procedure for trisecting angles, this 
procedure could be applied to the angle 60°. This would 
give us a construction for the angle 20°, since 60° itself 
is easily constructed. We shall prove the impossibility of 
constructing the angle 20° by Euclidean means. This is sufficient 
to show that no general procedure for trisecting angles 
can exist. 

The attack on the problem is algebraic. We translate 
geometrical constructions into algebraic terms — as is 
done in analytical geometry. It will be shown that any 
geometrical construction whatever is algebraically equivalent 

Trisection of an Angle 


to specifying a number with the help only of rational 
numbers and square root signs. For example, the number 

c = V2 + Vf + 5V3 

is specified by means of the rational numbers 2, -§, 5, 3, 
and some square root signs. This number corresponds to 
a geometrical construction. It is easy to construct geo- 
metrically two lines whose lengths are in the ratio c. 

If the angle 20° were constructible, it would be easy 
to draw lines having the ratio 2 cos 20°. Let w stand for 
the number 2 cos 20° for the remainder of this book. We 
shall show that w is the root of a cubic equation over the 
rationals, but not of any equation of lower degree. In 
fact, 1, w, w 2 are linearly independent over the rationals 
R; these numbers form a basis for R(w), and R(w) is of 
dimension 3 over R. 

If it were possible to construct 20° geometrically, it 
would be possible to represent w by an expression some- 
thing like the one that specifies c ; that is, an expression 
built up by means of square roots and rational numbers 

And here we have the germ of a contradiction. Any 
number specified by square roots, it can be shown, lies 
in a field of dimension 2 n over R, where n is a whole, 
positive number. Call this field F. 

Now we suppose w to lie in F. As F is a field, w 2 also 
must belong to F; thus 1, w, w 2 all belong to F, and any 
rational mixture of these also belongs to F. That is, R(w) 
is contained in F. F must have a dimension over R{w) 
(see page 202). It is easily seen that this dimension cannot 
be infinite. Call it q. Then the dimension of F over R is 
3 q (see page 205). 

Thus the presence of an element such as w> associated 
with a cubic equation, always betrays itself. If a field F 

210 A Concrete Approach to Abstract Algebra 

contains w, then the dimension of F over R must be 
divisible by 3. 

But any number specified by square roots lies in a 
field F of dimension 2” over R, and 2 n is never divisible 
by 3. 

Thus it is impossible for w to lie in any field of dimen- 
sion 2 n over R; it is therefore impossible for w to be ex- 
pressed by any collection of square roots and rational 
numbers. This means that the angle 20° cannot be con- 
structed by any Euclidean procedure. 

Three things, then, are required to fill out the details 
of this proof. (I) To show the correspondence between 
geometrical constructions and the repeated extraction 
of square roots. (II) To show that any number expressed 
by means of square roots lies in a field of dimension 2 n . 
(Ill) To show that w satisfies a cubic equation over the 
rationals R, but no simpler equation. We now consider 
these three in detail. 

I. Geometrical Constructions 

A certain amount of care is required to make sure that 
we do not overlook any possible type of geometrical 
construction. A construction might, for example, make 
use of a subsidiary construction; we might, say, construct 
a regular pentagon (which Euclid showed how to do) 
and then transfer an angle of 72° from this pentagon to 
some place in the main figure. There might be several 
such subsidiary constructions involved, and the figure 
might fall into several separate pieces. This is rather 
awkward for the argument we are using. However, we 
avoid this complication. If, as suggested above, we re- 
quired an angle of 72°, it would not matter where we 
constructed our regular pentagon, or how large its side 

Trisection of an Angle 


was. Accordingly, we could construct it on a line already 
marked in our main figure. This might be most incon- 
venient from the draftsman’s viewpoint; a very messy 
figure might result. Mathematically, it would make no 
difference at all. In this way, all the separate pieces of 
the figure could be joined into one connected figure. For 
the subsidiary constructions begin with some instruction 
such as, “Choose any two points” or “Draw any circle.” 
(If the subsidiary constructions did not begin in this way, 
they would have to mention specific points or distances, 
and the only specific points and distances are those 
already on the main figure, or connected to it in some 
way.) We are thus free to choose any points or any circle 
we like; we elect to choose points already on the diagram, 
or to draw a circle determined by the existing figure. 

In the same way, whenever an arbitrary element 
appears in a construction, we choose it to suit ourselves. 
For example, if A and B are known points, the perpen- 
dicular bisector of AB can be constructed by drawing a 
circle of arbitrary radius about A, and a circle of the 
same radius about B. The circles must of course be so 
chosen that they intersect. The particular radius used 
does not affect the result. So we are perfectly free to make 
the construction more definite: instead of an arbitrary, 
unknown radius, we select some suitable, known radius. 
In our example, we might use AB itself as radius. 

How do we know that there would always be some 
known length suitable for an arbitrary radius? In out- 
line, the argument is the following: a suitable circle is 
one that intersects a line or another circle; if the con- 
struction is possible at all, it means that some real num- 
ber will do for the radius of the circle; if l is the length 
of some known line, we can construct lines of length 
21 , 31, \l, and so on — in short, any rational multiple 


A Concrete Approach to Abstract Algebra 

of l — and thus get as near to any real number as we like. 
A sufficiently small change in the radius of a circle will 
leave it still intersecting in the required manner. 

Our argument will be in terms of analytical geometry. 
Suppose we set out in our attempt (foredoomed to 
failure) to construct an angle of 20°. In the course of 
our construction, we shall mark certain points. Let 0 
and P be the first two points that arise in the construe- 
tion. We take 0 as origin, and OP as unit distance along 
the x-axis. As the construction proceeds, we obtain 
points, lines, circles, all related in a definite manner to 0 
and P. Our earlier agreement cuts out all arbitrary 
elements. If we have to choose “any point” or “any 
distance,” we choose a point with rational coordinates, 
a distance that is a rational multiple of OP. Incidentally, 
Euclidean geometry allows us to construct the position 
of any point with rational coordinates; so we are not 
going outside Euclidea n construction when we select any 
such point. 

We are starting then with the points (0, 0) and (1, 0) 
marked. The permissible steps are: 

(i) To draw a line joining two known points. 

(ii) To mark the point where two known lines 

(iii) To draw a circle with known center and known 

(iv) To mark the points where a line cuts a circle. 

(v) To mark the points where circles intersect. 

(vi) To place the compass points on two known points, 
and then to move the compass without changing this 

All of these we want to consider in terms of analytic 
geometry. A point is specified by its coordinates (a, b ). 
A line is specified by its equation y = mx + c, unless it 
happens to be perpendicular to OP, when it will have 

Trisection of an Angle 


an equation x = k. A circle with center (a, b) and radius 
r has the equation (x — a) 2 + (y — b ) 2 = r 2 . 

In item (iii) we speak of “a known radius.” A radius 
is known when it is the distance between two points 
already determined. Item (vi) describes the operation 
performed when we are preparing to draw a circle with 
a prescribed radius. A radius, or a distance between two 
points, is of course specified by a single number, d. 

As we carry out operations (i) through (vi) successively, 
we obtain new geometrical objects at each step. These 
new objects — points, lines, circles, distances — are deter- 
mined by the old ones from which they arise. The num- 
bers specifying the new objects are functions of the 
numbers that have arisen earlier in the process. 

For example, suppose the figure, at some stage 
of construction, contains two lines with equations 
y = mix + fa, andy = m 2 x + fa. We can determine their 
point of intersection (provided they are not parallel). 
It is easily verified that the coordinates of the intersec- 
tion are rational functions of the numbers mi, m2, fa, fa. 

In operation (vi), if we have two points (a\, bi) and 
(a 2 , £2) in the figure, we can stretch the compasses from 
one point to the other. The distance, d, that now exists 
between the points of the compasses, is given by 

d = V(a 1 — a 2 ) 2 + ( bi — 6 2 ) 2 

Question: Verify that, in each of the operations (i) 

through (vi), the new numbers introduced arise from 
the old ones by addition, subtraction, multiplication, 
division, and extraction of square root, and never involve 
any other operation. 

Suppose now that any geometrical construction is 
given. As this construction is carried out, we keep a 
record of it in analytical form. Whenever a point is 
determined by the construction, we note down its coor- 


A Concrete Approach to Abstract Algebra 

dinates (a, b) . Whenever a line is drawn, we write down 
its equation, either in the form y = mx + c or in the 
form x = k, whichever is appropriate. We record all the 
distances, d , between points on the figure. When a circle 
is drawn, we note its equation. 

In this way, we arrive at a list of numbers. Some num- 
bers arise as coordinates, some as distances, some as 
coefficients in equations (like m and c for instance) . Each 
number is determined by numbers that occur earlier 
in the list. The earliest numbers in the list are rational 
numbers, 0 and 1. The only operations used in making 
the list are the rational operations — addition, subtrac- 
tion, multiplication, division — and the extraction of 
square roots. 

Thus, regardless of the way by which it enters into the 
geometrical construction, every number we meet is of 
the type stated earlier — formed with the help of rational 
numbers and square root operations alone. Statement 
(I) is thus justified. 

II. Fields of Dimension 2 n 

We now wish to show that every number in our list 
belongs to a field of dimension 2” over R, the rationals. 

Consider an example first of all. Suppose we construct, 
as is easy to do, the four points (0, 0), (1, 0), (1, 1), (1, 2). 
The lines y = 0, x = l, y = x, y = 2x join these points. 
The six distances between the four points are 1, 1, 1, 2, 
V 2, V5. (Circles must have been used to construct the 
right angle in the figure. In this example, we ignore these, 
to avoid complications.) 

The first numbers appearing above are rational; the 
last two, V 2 and V 5, are irrationals. 

Up to a certain stage, the numbers are included 
in R, the field of the rationals. When V2 appears, 

Trisection of an Angle 


we have to extend the field to i?(V 2), and when V5 
also appears, we have to make a further extension 
to /?(V 2, V5). 2) is of dimension 2 over /? 

(basis, 1 , V2); i?(V2, V5) is of dimension 4 over /?. 
The dimensions 2 and 4 are both powers of 2. This 
example illustrates the result we are hoping to prove. 

All the numbers listed for the construction above 
lie in the field i?(V 2, V5). We arrived at the field 
R(y/ 2, V 5) simply by adjoining to R all the irrational 
numbers in our list. We could, if we wished, adjoin the 
rational numbers on the list too, but that of course would 
make no difference, as they are already in R. 

It looks as though we have a great variety of cases to 
consider, for in other constructions we may meet not 
only expressions like V2 and V 5, but more complicated 

expressions like + 4'V / 5. However, we do not need 
to get entangled in these complications. There is essen- 
tially one situation that recurs again and again. 

Remember that we list the numbers in an order cor- 
responding to the order of geometrical construction. 
Each number is determined by the earlier numbers. We 
start with the rationals, R. As each new irrational num- 
ber occurs in the list, we adjoin it to the field. At each 
stage of the process, we thus have a field, which contains 
all the numbers in the list up to a certain point. Let F 
stand for the field obtained by adjoining to R the first 
m numbers on the list. Let h be the (m + l)-th number 
on the list, h is determined by the earlier numbers on the 
list; it can be expressed by a formula containing them, 
and this formula contains at most a single square root sign; 
no other irrational operation is involved. 

It may be that the formula giving h does not contain 
the square root symbol at all [for example, operations 
(i) and (ii)J. 


A Concrete Approach to Abstract Algebra 

Again, it may be that there is a square root operation 
in the formula, but that the numerical values allow this 
square root to be extracted without introducing any new 
irrationals. For example, the distance between two points 
involves a square root; but if the points happen to be 
the origin and the point (3, 4), the distance is given by 
the rational number 5, and no new irrational comes in. 
The same effect can occur with irrational expressions; 

the expression 59 + 30\/ 2 makes no difference if it is 
adjoined to i?(V 2), for 59 + 30 V 2 is the square of a 
quantity in i?(V 2), namely, 3 + 5V2. 

Here, then, is one possibility; the (m + l)-th number, 
h, may lie in the field F that contains the first m numbers. 
Then no action is called for, and we would pass on to 
consider the number after h in the list. 

The other possibility is that h is not in the field F. In 
this case, it must be of the form s + fV u, where s, t, u 
all belong to F; s, t, u stand for rational functions of the 
earlier numbers in the list; the earlier numbers all be- 
long to F, and F is a field; that is why s, t, u must belong 
to F. 

Accordingly, adjoining h to F is the same as adjoining 
Vu to F. u, of course, is not the square of an element 
of F. F(V u ) is thus the field obtained by adjoining the 
first {m + 1) numbers of the list to R. Suppose F has 
dimension p over R. Since F(V u) has dimension 2 over 
F, the dimension of F over R must be 2p (see chapter 10). 

Thus, at any such step, when we adjoin a number on 
the list, the dimension of our field is either unchanged or 
doubled. Since we start with the rational field, R, which 
is of dimension 1 over itself, we necessarily end up with 
a power of 2 as the dimension. This justifies state- 
ment (II). 

Trisection of an Angle 


III. The Cubic for w 

The formula cos 36 = 4 cos 3 6 — 3 cos 6 is a standard 
result in trigonometry. If we take the angle 20° for 6 
and write x = cos 20°, since cos 60° = 1/2, we have 
4x 3 — 3x — 1/2. This may be written 8* 3 — 6x — 1 = 0, 
or (2 x) 3 — 3{2x) — 1=0. This suggests bringing in a 
new symbol for 2x, and in fact w, defined earlier in this 
chapter as 2 cos 20°, is just that; w = 2x. Substitution 
gives the equation w 3 — 3w — 1=0. 

So w satisfies a cubic with rational coefficients, w 3 is 
accordingly a rational mixture of 1 and w, and every 
power of w is a mixture of 1 , w, w 2 . It looks as if R(w) is of 
dimension 3 over R, with 1, w, w 2 as the basis. Now this 
is in fact true, but unfortunately we cannot go straight 
to that conclusion. The reason why can be explained 
either in the language of vector spaces or of equations. 
We saw in chapter 10 that these were closely connected; 
we showed that a certain quantity satisfied an equation 
by considering its powers and proving them linearly 

The fact that w satisfies a cubic equation means that 
1, w, w 2 , w z are linearly dependent. But it may be that 
already 1, w, w 2 are linearly dependent; if so, R(w) will 
be of dimension 2 in general. It may even happen that 

1, w are linearly dependent; if so, w is rational and R(w) 
is the same as R; then R(w) has dimension 1. 

Thus the fact that w satisfies a cubic equation proves 
R(w) to be of dimension 3 or less. If R(w) is of dimension 

2, then w satisfies a quadratic equation with rational 
coefficients (see chapter 10). If R{w) is of dimension 1, 
then w satisfies a linear equation with rational coef- 

In terms of equations, it is easy to see how a quantity 


A Concrete Approach to Abstract Algebra 

can satisfy a cubic equation and also an equation of 
lower degree. For example, the cubic equation It 3 -\- 
6t — 13 = 0 holds for the rational number t = 1. The 
cubic equation s z + s 2 — 2s — 2 — 0 holds for the 
irrational quantity s = V2. This cubic is simply 
(j+1)(j*-2) =0. 

In both our examples, the cubic factors. We complete 
our proof by showing (i) that w could only satisfy an 
equation of lower degree if the cubic factored, and (ii) 
that the cubic x 3 — 3* — 1 does not factor into polyno- 
mials with rational coefficients. 

The proof of (i) follows a fairly natural train of 
thought. Let us take an example where a quantity 
satisfies two equations. There is an irrational number z 
that satisfies the two equations 

z 4 + 2z 3 + z 2 - 1 = 0, (I) 

z 3 + 3z 2 + z -2 = 0. (II) 

But from these two equations, we can derive yet another 
equation satisfied by z. If we subtract z times (II) from 
(I), we shall obtain an equation not containing any z 4 
term, namely, 

-z 3 + 2z - 1 = 0. (Ill) 

We now have two cubics, (II) and (III), satisfied by z. 
From these we can obtain an equation in which z 3 does 
not appear. In this particular example, we have simply 
to add (II) and (III). This gives 

3z 2 + 3z - 3 = 0 
or, on dividing by 3, 

+ z - 1 = 0. (IV) 

But when is this going to stop? We have now reached a 
quadratic; if we subtract z times (IV) from (II), we shall 

Trisection of an Angle 


obtain a quadratic, and by combining this with (IV), 
we should expect to obtain a linear equation for z. 
Actually, this does not happen. If we subtract z times 
(IV) from (II), we obtain 2z 2 + 2z — 2 = 0. Dividing 
this by 2, we obtain equation (IV) again. Equation (IV) 
is the simplest equation with rational coefficients satisfied 
by z. 

In effect, we have done a kind of long division above. > 
Combining the steps, equation (IV) results on sub- 
tracting (z — 1) times (II) from (I). In the language of 
division, when z 4 + 2 z 3 + z 2 — 1 is divided by z 3 + 

3 z 2 + z — 2, the quotient is ^ — 1 and the remainder 
is z 2 + z — 1. We are not particularly interested in the 
quotient. It is the remainder that turns up in equation 
(IV). The reason why the process cannot be carried any 
further is that z 2 + z — 1 divides exactly into the poly- 
nomials that occur in equations (I) and (II). It is in 
fact their H.C.F. You have probably noticed that the 
calculations above are those we should do to obtain the 
H.C.F. of polynomials (I) and (II) by the method of 
chapter 4. 

The object of these considerations is to show that 
H.C.F. comes quite naturally into the picture. The work 
above suggests that if a quantity z is a root of the equa- 
tion f(x) = 0 and g(x) — 0, then it is also a root of the 
equation h(x) = 0, where h{x) is the H.C.F. of /(z) 
and g(z). 

It should not be hard to prove this, because we have 
only a few theorems about H.C.F. (The fewer theorems 
there are, the more quickly one can look through them 
and decide which are relevant. This may sound a para- 
dox, but it is not.) The statements (i), (ii), (iii) of page 
88 pretty well sum up what we know about H.C.F. 
Statement (ii) is the one that will help us. The form of 

220 A Concrete Approach to Abstract Algebra 

it, appropriate to polynomials, is the following; if h(x ) is 
the H.C.F. of /(x) and g(x), then there exist polynomials 
a(x), b(x) such that 

h(x) = a(x)f(x) + b(x)g(x). (V) 

The desired result follows immediately, z is a root of 
/(x) == 0; this means f(z) = 0. z is a root of g(x) = 0; 
so g(z) = 0. Substitute z for x in equation (V). The 
right-hand side clearly becomes zero, as each term con- 
tains a zero factor. Hence h(z) = 0. This is what we 
wanted to show; z is a root of the polynomial h{x). 

Equation (V) also shows that the common factor A(x) 
cannot be a trivial one. For example, if /(x) were x 3 — 1 
and g(x) were x 2 + 1, the H.C.F. h(x ) would simply be 1. 
The two polynomials have only a trivial common factor 
if that factor is constant. But this cannot happen in the situa- 
tion that interests us. For suppose it did. Then h{x) is 1 
and equation (V) becomes 

1 = a(x)f(x) + b(x)g(x). 

We suppose that/(x) and g(x) have the common root x. 
Substitute z for x. The right-hand side becomes zero, 
and we have 1=0. But this is impossible. 

Accordingly, if two polynomials with rational coeffi- 
cients, /(x) and g(x), have a common irrational root z, 
this quantity is also a root of their H.C.F., h(x); and h(x) 
is not trivial, not constant, but a genuine, healthy poly- 
nomial. h(x) can be calculated by the H.C.F. process, so 
that h(x) also is a polynomial with rational coefficients. 

Now we come to apply this to our quantity w. We 
know that w is a root of the polynomial x 3 — 3x — 1. 
We choose this polynomial for/(x). We are investigating 
whether w can satisfy also a quadratic or linear equation. 
If w is the root of a linear or quadratic polynomial, call 
that polynomial ^(x). Then, by the argument above, 

Trisection of an Angle 


there is a polynomial h(x) that is a factor both of f(x ) 
and g(x), and this h(x) is not just a constant. Since h(x) 
is a factor ofg(x), the polynomial h(x) is either linear or 
quadratic — for g(x) is supposed quadratic at most. Since 
h(x) is a factor of f(x ), it follows that f(x) can be split 
into rational factors. All of this, of course, is on the 
assumption that w satisfies some equation of the first 
or second degree — an assumption we are trying to dis- 
prove by showing that it leads to a contradiction. 

We have now established our contention (i); if w 
satisfies any equation simpler than the cubic, then the 
cubic factors. 

We finally establish the contradiction by proving 
assertion (ii) ; the cubic cannot be factored into polyno- 
mials with rational coefficients. 

This proof is quite long enough already, and we shall 
shorten this final stage by quoting a standard theorem 
without proof. The line of argument followed is one 
that — in a loose form — would occur to any competent 
high school student of algebra. 

Suppose a student is asked to find out whether 
2x 2 + x + 3 can be factored. The student will consider 
the possibility of such factors as x + 1, * + 3, 2* + 1, 
2x + 3, perhaps also x — 1, x — 3, 2x — 1, 2x — 3. 
That is, he considers whole number coefficients only, 
and these he chooses in a special way. He only considers 
for the constant term whole numbers that are factors 
of 3, and for the coefficient of x he only considers fac- 
tors of 2. He does not consider the possibility, say 
(1* + f) (3* + !)• In so doing, he is right (though he 
does not know why he is right) . It was proved by Gauss 
that, if we have a polynomial with whole number coeffi- 
cients, and we cannot factor it by means of polynomials 
with whole number coefficients, then we cannot factor 
it by bringing in fractions either. If such a polynomial 


A Concrete Approach to Abstract Algebra 

does not factor with the help of whole numbers, then it 
is irreducible over the rationals. 

Now it is very easy to see that x s — 3x — 1 cannot be 
factored into polynomials with whole number coeffi- 
cients. If it could be factored as (ax + b)(cx 2 + dx + e), 
we should have ac = 1 and be = — 1 . That is, a would be 
a factor of 1 and b a factor of — 1 . This restricts a and b 
to the values 1 and —1, if a and b are whole numbers. 
Thus x + 1 and x — \ are the only possible factors. It 
is easily seen that neither of these works. Thus the cubic 
cannot be factored with whole number coefficients; by 
Gauss’ result, it cannot be factored into polynomials 
with rational coefficients. Assertion (ii) is now justified. 
We have gone round plugging one leak after another, 
and we hope the argument is by now reasonably water- 

The Gauss theorem that we have quoted can be 
proved by an elementary proof of an essentially arith- 
metical nature. The argument we have given shows how 
modern algebraic concepts — fields, vector spaces, and 
their dimensions — can be used to prove theorems about 
elementary geometry and algebra. 

Answers to Exercises 

(PAGES 15-16) 

1. Yes; the classes a + b and ab are determined. 















+ I 




X I 












(1) through (7) do apply. 

2. Again, the classes are determined. (1) through (7) apply. 


















+ II 





















































3. The tables for 7, 11, 13 

• • • resemble those found in ques- 

tions 2 and 3. But for 4, 

6, 8, 

9, 10, 12 •• 

• the multiplication 

tables differ from those in question 2 and 3. For instance, with 

the number 6, we have the II times table 


Answers to Exercises 

II- 0 = 0, II- I = II, II -II = IV, 

II -III = O, II -IV = II, II-V = IV. 

The products O, II, IV each occur twice. But in questions 2 
and 3, in each multiplication table every symbol occurs once 
only. 3, 5, 7, 11, 13 are prime numbers. 4, 6, 8, 9, 10, 12 can 
be factored. This is the reason for the difference. 




































+ 4 






X 4 






























(1) through 

(7) do 


Are these tables isomorphic with those in question 2? The 
addition tables suggest that we let O correspond to 0, I to 2, 

II to 4, III to 6, IV to 8. But this does not work with the mul- 
tiplication table. I • I = I would correspond to 2 • 2 = 2, which 
is not so. We are tempted to leap to the conclusion that the 
systems are not isomorphic. But such a conclusion is only 
justified if we can show that there is no way of establishing a 
correspondence. We have only tried one way so far. We try 
again. Consider property (7). In the system above, which ele- 
ment plays the role of I? That is, which one makes no differ- 
ence when you multiply by it? Answer, 6. Try I corresponding 
to 6. Then II = I + I would have to correspond to 6 + 6 = 2. 

III = II + I would correspond to 2 + 6 = 8, and IV = 
III + I to 8 + 6 = 4. O corresponds to 0. You can now check 
that any result in question II, “translated” by this scheme, 
gives a correct result in the 0, 2, 4, 6, 8 tables. The systems are 


II- II = IV II- II- II = HI II* II* II* II = I 

IV. IV = I IV- IV- IV = IV IV- IV- IV- IV = I 

Answers to Exercises 


One can of course calculate higher powers than these. The 
answers form a repeating pattern. 

6. Subtraction is possible in all the arithmetics considered 
in questions 1 through 3. For the arithmetics corresponding to 
prime numbers, division with a unique answer exists (division 
by 0 excluded, of course). In the non-prime cases, one either 
gets no answer or several answers. Thus, in the arithmetic 
arising from 6, II -x = III has no solution, but II-x = IV has 
two solutions, x = II and x = V. 

7. The perfect squares are O, I, IV. No number is prime. 
Even I has factors I = II • III = IV- IV. The arithmetic has 
no need of negative numbers and fractions. Subtraction and 
division can always be carried out, without introducing any 
new symbols. 

(PAGE 20) 

Each symbol occurs exactly once in each row of the addition 



1. Number 



Fourth Power 





















(Compare question 5, page 16.) 

By inspection of the tables above, x 2 = 1 for x = 1 and x = 4; 
x 3 = 1 only for x = 1 ; x 4 = 1 for x = 1, 2, 3, 4. 

2. As in ordinary algebra, we find 

(x + y) • (x + y) • (* + y) = x 3 + 3 x 2 y + 3 xy 2 + y 3 . 

As 3 + 3 = 1, modulo 5, the next multiplication by x + y gives 

(* + >)•(*+>)•(* + ?)•(*+ y) 

= x 4 + 4 x 3 y + x 2 y 2 + 4 xy 3 + y 4 . 


Answers to Exercises 

The final multiplication, since 4+1 = 0 modulo 5, gives 
(x + y) • (x + y) ■ (x + y) • (x + y) • (x + y) = + + y 5 . 

Many points for discussion arise from this question. Can 
you have a 5th power when working modulo 5, since only the 
numbers 0, 1, 2, 3, 4 are admitted? See page 57. 

3. (+ + l)/(x + 2) = x + 3 modulo 5. This answer may 
also be written * — 2. x 2 + 1 = 0 for x = 2, 3. 

4. x 2 + x + 3 = (x + 2)(x + 4). 

5. (i) 1, 2; (ii) 1, 1; (iii) 2, 2; (iv) 1, 4. 

(i) (x - l)(x - 2); (ii) (x - l) 2 ; (iii) (x - 2) 2 ; 

(iv) (x - l)(x - 4). 

6. Quotient x 2 + 4x + 1, remainder 1. 

7. Yes. 

8. Yes. x = 1 is a solution. 

9. Values are 2, 0, 0, 2, 0, 0. The equation has four roots, 
1,2, 4, 5. 

10. (i) All except (70) and (72). 

(ii) All except (70) and (72). 

11. (i) Clearly not. Question 9 gives an example where a 
quadratic has four roots, although — see question 10 (i) — state- 
ments (7) through (9) and (77) hold. 

(ii) Yes. The quadratic with roots a, b is (x — a)(x — b) 
= 0. If a third root c, different from a and b existed, we should 
have (c — a)- (c — b) = 0. Now use statement (72). 

12. (i) Yes. x 2 + 2 for example. 

(ii) Yes. x 2 + 2 for example. 

13. It has no other solution. (Compare question 1.) 
(a 3 — 22) /(x — 3) = x 2 + 3x + 4. This quadratic has no fac- 
tors. If it had, we should be able to find further solutions of 
+ -2 = 0 . 

(PAGE 31) 

1. All but (8), (70), (77). 

2. All but (70), (77). 

3. All tests passed, a field. 

4. A field. 

Answers to Exercises 


5. A field. 

6. All but (70). 

7. Field. 

8. Passes (7), (3), (5), (8), (9) only. Perhaps (77). 

9. All but (70). 

10. Field. 

11. Field. 

12. Field. 

13. All but (70), (72). 

(PAGES 31-32) 

Question for investigation. This question is answered in chap- 
ter 3. 

(PAGE 36) 

Modulo 3. x(x — l)(x — 2) — x 3 — x. 

Modulo 5, x(x — l)(x — 2)(x — 3)(x — 4) = x 5 — x. 

(PAGE 70) 

1. When the polynomial f(x) is divided by the polynomial 
x — a, let the quotient be the polynomial q(x) and the re- 
mainder the constant polynomial R. Then 

f(x) = (x — a) • q(x) + R • • • (I). x indeterminate. 

By the theorems in chapter 2, if, in equation I, the indeter- 
minate x is replaced by any fixed element of F, we obtain a 
true statement. Suppose the indeterminate x is replaced by the 
fixed element a. We then have 

f(a) =0 -q(a) + R'- - (II). 

Equation II is a statement about fixed elements of F. 

Now Q'q(a) = 0 and 0 + R = R. 

Hence /(a) = /?••• (III). 

This is the remainder theorem. Note that R is used in two 
Senses. In equations II, III, it stands for a particular element of 


Answers to Exercises 

F; in equation I it denotes the constant polynomial, correspond- 
ing to that element. 

2. By saying that the polynomial ax 2 -}- bx + c has at most 
two roots, we mean that it is impossible to find three distinct 
elements k, m, n of F such that 

ah 2 + bk + c = 0 • • • (I) 

am 2 + bm + c = 0 ■ • • (II) 

an 2 + bn + c = 0 • • • (III) 

We assume that a is not zero. 

Suppose that the three equations above were satisfied. With 
the help of the remainder theorem, equation I shows that the 
remainder would be zero if ax 2 + bx + c were divided by 
x — k. That is, x — k is a factor of ax 2 + bx + c. Similarly, 
from equation II, x — m is a factor of ax 2 + bx + c. 

Hence ax 2 + bx + c = a(x — k)(x — m). In this equation, 
x is an indeterminate. By the statement of chapter 2, a true 
statement results if x is replaced everywhere by the fixed ele- 
ment n. 

Hence an 2 -f* bn -f- c = a(n — k)(n — m ). 

Using equation III, 0 = a(n — k)(n — m). 

You can now show that this contradicts field axiom (72). 
For we have assumed a ^ 0. b — k ^ 0 and n w ^ 0 since 
n, k, m were supposed distinct. 

(PAGE 80) 

In question 1, a solution is given. Others exist. 

1 (i). x = —3,y = 2. 

1 (it), x = 5, y — — 3. 

1 (Hi) . x = 5, y = — 7. 

2. No. For all integers *, y, the expression 4x + 6y gives 
an even number; it can never give the odd number 1 . 

3. k must be divisible by h, the H.G.F. of a and b. 

(PAGE 81) 

1 . lx + 17y = 1 has a solution x = 5, y = — 2 . So 
7.5 _ 17.2 = 1 . Hence 7-5 = 1 modulo 17. So 5 is the 
reciprocal of 7. 

Answers to Exercises 


As 3*11 — 2-17 = — 1, the reciprocal of 11 is —3, which 
equals 14 modulo 17. 

2. 6 and 241. 

3. 2/7 = 5, 5/6 = 10. 

(PAGE 84) 

Question for discussion. No. They satisfy the same axioms, 
but it is not possible to establish a correspondence between 
the elements, such as is required for isomorphism. The integer 1 
and the constant polynomial 1 must correspond. To preserve 
the addition tables, the integer 2 = 1 + 1 must correspond 
to the polynomial 2; repeating the argument, the integer n 
must correspond to the constant polynomial n, for every posi- 
tive n. It is easily seen that this must hold also for negative n. 
Now we have used up all the integers, and have none left to 
put into correspondence with x, x 2 , etc. 

(PAGES 86-89) 

Exercises left to reader. 

(PAGES 106-107) 

1. If x 2 + 1 had a factor x — a, by the remainder theorem 
a 2 + 1 would be zero. Try a = 0, 1, 2 in turn. The theory can 
be constructed. 

2. x 2 + 3 irreducible by argument similar to that in ques- 
tion 1. x 2 + 1 = (x — 2) (* — 3), so reducible. 

Procedure (ii) gives us complex numbers modulo 5. 

Procedure (i) may be interpreted in two ways. If we regard 
a + by/'— 1 as giving 25 distinct elements, we find for example 
(2 + V — 1) • (2 — V — 1) = 0 and axiom (12) fails. On the 
other hand, if we extract the square root, and write say 
V — 1 = 2 modulo 5, we merely obtain the arithmetic mod- 
ulo 5 again, described in a wasteful manner. In this sense, we 
obtain a field, but not a new one. 


Answers to Exercises 

3. x-x = x 2 ; x(x +1) = x 2 -f- x; (x -j- 1 ) (at + 1) = x 2 + 1. 
The quadratic x 2 + x + 1 does not appear in this list. It is 
irreducible, and is the only irreducible quadratic. 




M + I 










M + I 




M + I 


M + I 


M + I 






M + I 










M + I 





M + I 

M + I 


M + I 

M + I 


Axioms (70) and (72) fail. 

6. Yes. 

7. (2 + 3Q) • (4 + 5Q) = 8 + 22Q. 

No. For example, Q-Q = 0, so axiom (72) fails. The division 
1/Q is impossible, so axiom (70) fails. 

8. f(x) = (a — bx)/(a 2 + b 2 ). 

k = b 2 / (a 2 + b 2 ). 

From/(x), we obtain the reciprocal 

(a - bJ)/(a 2 + b 2 ). 

9. Nothing new, just the reals again. The graphs intersect 
on the y-axis. 

10. x(x — l)(x — M)(x — M — 1) = x* — x. So x 4 = x is 
the required equation. 

(PAGE 121) 

Addition, subtraction, and multiplication give no difficulty. To 
establish division, we show that each element, except zero, has 
a reciprocal. In fact, (rV 7 4 + xV / 2 t)(a^/ A + i V 7 2 + c) = 1 

if a = (s 2 — rt)/k, and b = (2r 2 — st)/k, and c = (t 2 — 2 rs)/k 
where k = 4r® + 2 j 3 + t 3 — 6rst. (We ought to show that 

Answers to Exercises 


k is never zero for rational r, s, t. To do this requires a 
rather expert knowledge of elementary algebra. One might 
make use of the well-known factoring x 3 y 3 + z 3 — 3 xyz = 
I {* + y + z) {(* — y ) 2 + (y — z) 2 + (z — x ) 2 } and put 
x — r^A, y = s^l, z = t. It is a merit of modern algebra that 
it avoids this troublesome detail.) 

The reciprocal of + 2'V / 2 -f- 3 is (v^ — 4v / 2 + 5)/ll. 

(PAGE 129) 

1. n = 7. 

2. All the non-zero elements except 1 have this property. 

3. Q, Q 2 , Q 4 are all roots. 

4. Q 3 , Q 5 , Q 6 are roots of x 3 + x 2 + 1 = 0. By the re- 
mainder theorem, x 3 + x 2 + 1 = (x — Q 3 )(x — Q 5 )(x — Q 6 ), 
hence reducible. 

5. If we introduce R, satisfying R 3 + R 2 + 1 = 0, we again 
get a field with 8 elements. An isomorphism can be established 
by making R correspond to Q 3 , R 2 to Q 6 , and so on. (There are 
also two other ways of establishing an isomorphism.) The 
i?-field is thus the Q-field in disguise. 

6- Q , Q 2 , Q 4 - Exponents are powers of 2. Note Q 8 = Q, 

Q 16 = Q 2 , • • • 

(PAGE 147) 

1. S = P + Q, T = Q - P. 

2. P=|(.S-n Q = i(S+T). 

3. — . 

4. (a) Yes. (b) Yes. (c) No. (i) fails, (d) No. (ii) fails. 

5. (a) No. (i) fails, (b) No. (ii) fails. 

6. U, V form basis if ad — be 0. 

(PAGES 161-162) 

1-5 ask for proofs. 

6. (i) V.8. (ii) V.7 and V.9. 


Answers to Exercises 

(PAGE 166) 

1. Linearly independent. 

2. Linearly dependent. 9A B — 28 C = 0. 

3. Linearly dependent. A + B + C = 0. 

4. Linearly independent. 

(PAGE 174) 

Question, x, y, z. 

(PAGE 177) 

Answers different from those below are not necessarily wrong. 
A system can be reduced to standard form in many ways. 

1. * + y + z; y + z + t; z + t + u. 

2. * + 2y + 3z + At; y + 2z + 3t. C = 2B - A. D = 
3 B - 2 A. 

3. x — 2y + z, y — z. C = —A — B. 

4. x — y, y — z. C — —A — B. 

5. x + y, y + z, z. 

(PAGE 194) 

1 . — . 

2. They form a vector space of four dimensions. The equa- 
tion is of the fourth degree. 

3. They form a basis, m 4 — 10m 2 +1=0. The roots of 
this equation are ±V2 rhV 3. 

4. 1, r, r 2 • • • r n_1 form basis, n dimensions. Degree n. 

(PAGE 198) 

Fourth degree, x 4 — 10* 2 +1 = 0. 

Answers to Exercises 


(PAGE 202) 

1 . 2 . 

2. Case (i). 

3. 3. 

4. 2. 

5. 2. 

6. 2, 3. 

This brief, understandable introduction to abstract 
algebra will appeal to students, teachers, and others who 
want to know more about modern mathematics. Ab- 
stract algebra is a creation of this century. Three hun- 
dred years of mathematical research separate it from the 
traditional high school mathematics, most of which was 
known in the seventeenth century. It strikes many of us 
as something strange, unfamiliar, mysterious. Even when 
a mathematician takes pains to explain modern algebra 
clearly, he too often remains incomprehensible to his 
audience. In order to bridge this gap between the solid 
ground of traditional algebra and the abstract territory 
of modern algebra, the author has worked out a very 
carefully devised progression from the concrete to the 
abstract, from the known to the unknown. Each ques- 
tion is posed concretely: examples are worked, evidence 
is collected, and the reader is led along a line of thought 
that may enable him to discover a result before it is 
actually stated in the book. Those points that may con- 
fuse the beginner arc discussed very carefully. 

W. W. Sawyer, professor of mathematics at Wesleyan 
University, is a graduate of St. John’s College, Cam- 
bridge, England, where he specialized in relativity and 
quantum theory. His teaching career carried him from 
England to West Africa to New Zealand. In 1956, he 
was invited to the United States to contribute to mathe- 
matical education. Professor Sawyer is the author of 
many articles and several books (including the very 
popular Mathematicians’ Delight and Prelude to Mathe- 
matics). He also edits The Mathematics Student Journal , 
a magazine for high school students, and is well known 
as a speaker at teachers’ conferences. 


San Francisco 

Concrete Approach to Abstract Algebra